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You can choose f(x) sufficiently close to A, say A + ϵ. Then it is possible to find a δ, such that if x is
within δ of a, then f(x) will be within ϵ of A.
f(x)
δ(ϵ) > 0
x → a A = lim f(x)
f(x)
x→a
ϵ>0
0 < |x − a| < δ
| f(x) − A | < ϵ
The number A is called the limit of a function f(x) as x → + ∞, A = lim f(x), if for any ϵ > 0
x→+∞
there exists a number M(ϵ) > 0 and belonging to the domain of the definition of the function f(x) the
inequality | f(x) − A | < ϵ holds true.
Example: Prove lim(3x − 8) = − 5
x→1
We need to prove that for any ϵ > 0 there exist δ > 0 such that from the inequality | x − 1 | < δ it
follows that f(x) − (−5) = f(x) + 5 < ϵ.
In other words we need to solve the inequality
3x − 8 + 5 = 3 x − 1 < ϵ. ⟹
ϵ
=δ
x−1 <
3
You can get f(x) as close as you want to A by making x sufficiently close to a.
You can choose f(x) sufficiently close to A, say A + ϵ. Then it is possible to find a δ, such that if x is
within δ of a, then f(x) will be within ϵ of A.
Ashani Ray Classes (ashaniray.classes@gmail.com)
lim f(x) = A
x→a
x1, x2, . . . , xn, . . .
(belonging to the domain of definition of the function and differing from a) the corresponding sequence
of values of y,
y1 = f(x1), y2 = f(x2), …, yn = f(xn) ,…
has a limit, which is the number A
The definition of the limit after Heine is conveniently applied when we have to prove that a function
f(x) has no limit. For this it is sufficient to show that there exist two sequences {xn′} and {xn′′} such
that lim x′ = lim x′′ = a but the corresponding sequences {f(xn′)} and {f(xn′′)} do not have
n→∞
n→∞
identical limits.









Ashani Ray Classes (ashaniray.classes@gmail.com)
a) lim sin
x→1
1
x−1
Solution: Choose two sequences xn = 1 +
For which lim x = lim x′ = 1
n→∞
1
2
and xn′ = 1 +
, (n=1,2,…)
nπ
(4n + 1)π
n→∞
The corresponding sequences of values of the function are:
f(xn) = sin
(1 +
1
1
nπ )
−1
= sin nπ = 0
π
4n + 1
=1
f(xn′) = sin
= sin
π = sin 2nπ +
(
)
2
2
2
(1 + (4n + 1)π ) − 1
Hence lim f(xn) = 0 and lim f(xn′) = 1.
xn→1
1
xn′→1
i.e. the sequences {f(xn′)} and {f(xn′′)} have different limits, whence it follows that lim sin
x→1
does not exist.








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1
x−1
lim u(x)
lim v(x)
x→a
x→a
1. lim [u(x) ± v(x)] = lim u(x) ± lim v(x)
x→a
x→a
x→a
2. lim [u(x) ⋅ v(x)] = lim u(x) ⋅ lim v(x)
x→a
x→a
x→a
limx→a u(x)
u(x)
=
3. lim
, lim v(x) ≠ 0
[
]
(x→a
)
x→a v(x)
limx→a v(x)
v(x)
4. lim [u(x)] = lim u(x)limx→a v(x)
x→a
x→a
Ashani Ray Classes (ashaniray.classes@gmail.com)
sin x
lim
=1
x→0 x
x
1
1
lim 1 +
= lim (1 + α) α = e
x→∞ (
x)
α→0
ln(1 + x)
lim
=1
x→0
x
ex − 1
lim
=1
x→0
x
Note: If lim [ f(x)]
ϕ(x)
x→a
lim [ f(x)]
ϕ(x)
x→a
= lim e
x→a
is of the form 1∞ then the following method is useful:
= lim [1 + {f(x) − 1}]
ϕ(x)
x→a
ϕ(x)[ f(x)−1]
= e limx→a ϕ(x)[ f(x)−1]
= lim {1 + ( f(x) − 1)}
x→a [
Note: a = e ln a. Hence a x = e x ln a
Ashani Ray Classes (ashaniray.classes@gmail.com)
1
f(x) − 1
]
ϕ(x)[ f(x)−1]
4x 5 + 9x + 7
lim 6
x→1 3x + x 3 + 1
15
x 3 + 3x 2 − 9x − 2
2. lim
[ Ans: ]
x→2
x3 − x − 6
11
x+1
3. lim
[Ans: 1]
x→−1
2
6x + 3 + 3x
4.
2
x3
x2
lim
−
[Ans: ]
9
x→+∞ ( 3x 2 − 4
3x + 2 )
5.
lim ( 9x 2 + 1 − 3x)
x→+∞
2 x+3 x+5 x
3
6.
7.
8.
9.
lim
x→+∞
3x − 2 +
[Ans: 0]
5
3
2x − 3
[ Ans:
lim ( 2x 2 − 3 − 5x) [Ans: +∞]
x→−∞
lim
x→+∞
2
3
]
2
2
2x 2 + 3
2x 2 + 3
and lim
[Ans:
and −
]
4
4
4x + 2
x→−∞ 4x + 2
2x
lim 5 x + 3 [Ans: 25]
x→+∞
Ashani Ray Classes (ashaniray.classes@gmail.com)
lim
x→1
11. lim
x→0
2x − 2
3
26 + x − 3
k
1+x−1
x
1
, k is positive integer [Ans: ]
k
sin (x − 6 )
π
12. limπ
x→ 6
[Ans: 1]
3 − 2 cos x
1 − cos x
1
[Ans: ]
x→0
x2
2
1
tan x − sin x
14. lim
[Ans: ]
2
x→0
x3
2
13. lim
8x +3
2x + 3
x→∞ ( 2x 2 + 5 )
15. lim
2
e x − e −x
16. lim
x→0 sin x
1+x
x→1 ( 2 + x )
17. lim
(1 − x)
(1 − x)
[Ans:
2
]
3
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Ashani Ray Classes (ashaniray.classes@gmail.com)
2 x+3 x+5 x
3
lim
3x − 2 +
x→+∞
Note:
(2x − 3)
x
1
2
1
3
=
2+
5
3
= lim
2x − 3
(2x − 3)
3
x6
x→+∞
2
6
3−
2
x
+
(2x − 3)2
=
x3
(
)
3
6
x
6
+
4
x
5
10
−
x3
12
x2
+
9
x3
=
2
3
1
6
lim ( 2x 2 − 3 − 5x) = lim ( 2x 2 − 3 + (−5x)) = ∞
x→−∞
x→−∞
lim
x→+∞
lim
x→−∞
x
2
2+
3
x2
2+
2
2x + 3
= lim
= lim
=
2
2
x→+∞
x→+∞
4
4x + 2
x (4 + x )
(4 + x )
−x
2
2+
3
x2
2+
3
x2
2
2x + 3
= lim
= lim −
=−
2
2
x→−∞
x→−∞
4
4x + 2
x (4 + x )
(4 + x )
2x
2x
lim 5 x + 3 = 5limx→+∞ x + 3 = 52 = 25
x→+∞
3
x2
3
x
2
x
1
1
1
= x 3 − 2 = x− 6 =
1
6
x
lim
26 + x = z 3
2x − 2
x→1
3
x = z 3 − 26
x→1
z→3
2z 3 − 54
2(z − 3)(z 2 + 3z + 9)
= lim
= lim
= lim 2(z 2 + 3z + 9) = 54
z→3
z−3
z→3
26 + x − 3 z→3 z − 3
11. Substitute 1 + x = z k Therefore x = z k − 1 and z → 1 as x → 0. Therefore
k
1+x−1
z−1
= lim k
lim
z→1 z − 1
x→0
x
π
π
π
12. Substitute z = x − . Then x = z + , and z → 0 as x →
6
6
6
limπ
sin (x − π6 )
x→ 6
= lim
z→0
3 − 2 cos x
2 sin
2 3
z
sin2 2
sin z
= lim
3 − 2 cos (z + π6 )
z→0
z
2
cos
z
2
+ 2 sin
z
2
2 sin2
1 − cos x
lim
= lim
2
x→0
x2
x→0
x
cos
x
2
z
2
= lim
= lim
x→0
x2
24
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z→0
cos
x
2
z
2
sin z
3−
z
2
+ cos
z
2
sin
1
= lim x
2 x→0 2
x
2
3 sin
z→0
sin2
= lim
3 cos z + sin z
=1
⋅
sin
x
2
x
2
=
1
2
tan x − sin x
sin x(1 − cos x)
1
sin x 1 − cos x 1
=
=
lim
=
lim
⋅
⋅
3
3
2
2
x→0
x
x→0
x cos x
x→0 cos x
x
x
2
2x 2 + 3
2x 2 + 3
−2
f(x) − 1 = − 2
f(x) = 2
=
1
+
2x + 5
2x + 5
2x 2 + 5
2x 2 + 5
lim
8x 2+3
2x 2 + 3
2
2x + 3
lim
(8x
+3)
limx→∞(8x 2+3)[− 22 ]
x→∞
limx→∞ ϕ(x)[ f(x)−1]
2 + 5 − 1]
[
2x
2x + 5 = e −8
=e
lim
=e
=e
2
x→∞ ( 2x + 5 )
e x − e −x
e x − 1 − e −x + 1
ex − 1
e −x − 1
lim
= lim
= lim
− lim
x→0 sin x
x→0
sin x
x→0 sin x
x→0 sin x
x
−x
e −1
x
e −1 x
=2
= lim
⋅
+ lim
x→0
x
sin x x→0 −x sin x
e −x (e 2x − 1)
e −x (e 2x − 1)
e x − e −x
x
Alternate: lim
= lim
= lim
⋅
⋅2
x→0 sin x
x→0
sin x
x→0
2x
sin x
2x
h
e
−
1
e
− 1)
x
(
)
(
−x
= 2 ⋅ lim e ⋅ lim
⋅ lim
= 2 ⋅ 1 ⋅ lim
⋅ 1[ Substituting h = 2x, when x → 0
x→0
x→0
2x
x→0 sin x
h→0
h
then h or 2x → 0] = 2
2
1
1
1
sin x
= lim t ⋅ sin . The function sin is bounded between +1 and −1 making lim t ⋅ sin = 0
t→0
t
t
t→0
t
x→∞ x
lim
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lim
x→0
tan x − sin x
sin x(1 − cos x)
1
sin x 1 − cos x 1
=
=
lim
=
lim
⋅
⋅
3
3
2
2
x
x→0
x cos x
x→0 cos x
x
x
2x 2 + 3
−2
=1+ 2
2x 2 + 5
2x + 5
2
8x +3
2x 2 + 3
lim
x→∞ ( 2x 2 + 5 )
=e
limx→∞(8x 2+3)[−
= lim
x→∞
2
2x2 + 5 ]
−2
1+ 2
(
2x + 5 )
= e −8
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2x 2 + 5
−2
−2
⋅8x 2+3
2x2 + 5
= lim e
x→∞
−2
⋅8x 2+3
2x2 + 5
α(x)
x→a
x → ∞ lim α(x) = 0
x→a
lim α(x) = 0
x→∞
sin x
x→∞
x
A function f(x) is called infinite as x → a or x → ∞ if lim f(x) = ∞ or lim f(x) = ∞
α(x) =
x→a
x→∞
If the function α(x) and β(x) are infinitesimals as x → a and if α(x) ∼ γ(x), β(x) ∼ δ(x), then
lim
x→a
α(x)
γ(x)
= lim
[Replacing an infinitesimal by an equivalent one]
β(x) x→a δ(x)
For two infinitesimal functions to be equivalent it is necessary and sufficient that their difference be an
infinitesimal of.a higher order as compared with each of the two.
The following are equivalent infinitesimal functions:
• sin x ∼ x
• tan x ∼ x
x2
• 1 − cos x ∼ 2
• sin−1 x ∼ x
• tan−1(x) ∼ x
• ln[1 + x] ∼ x
• a x − 1 ∼ x ln a (a>0, in particular, e x − 1 ∼ x)
x
P
1
+
x
−
1
∼
[1
+
x]
−
1
∼
Px
,
in
particular
•
n
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n
• sin x ∼ x
• tan x ∼ x
x2
• 1 − cos x ∼ 2
• sin−1 x ∼ x
• tan−1(x) ∼ x
• ln[1 + x] ∼ x
• a x − 1 ∼ x ln a (a>0, in particular, e x − 1 ∼ x)
x
P
1
+
x
−
1
∼
,
in
particular
[1
+
x]
−
1
∼
Px
•
n
n
Ashani Ray Classes (ashaniray.classes@gmail.com)
0 ∞
0 ∞
If the function f(x) and g(x) are differentiable in a certain neighbourhood of the point a, except may be at the
point a itself, and g′(x) ≠ 0 and if lim f(x) = lim g(x) = 0 or lim f(x) = lim g(x) = ∞ , then
lim
x→a
x→a
f(x)
f ′(x)
= lim
g(x) x→a g′(x)
x→a
x→a
x→a
f ′(x)
provided the limit lim
exists (L’Hospitals’s rule)
x→a g′(x)
xn − an
Example: Evaluate lim
x→a x − a
Solution: lim x n − a n = lim x − a = 0. Also both (x n − a n) and (x − a) are differentiable in the
x→a
neighbourhood of a.
x→a
nx n−1
xn − an
= na n−1
= lim
Therefore lim
x→a
1
x→a x − a





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x2
xn
e =1+x+
+ ... +
+ o(x n)
2!
n!
2n−1
x3 x5 x7
n−1 x
sin x = x −
+
−
+ . . . + (−1)
+ o(x 2n)
3! 5! 7!
(2n − 1)!
2n
x2 x4 x6
n x
cos x = 1 −
+
−
+ . . . + (−1)
+ o(x 2n)
2! 4! 6!
(2n)!
x3
2 2
tan x = x +
+
x + ...
3
15
α(α − 1) 2
α(α − 1) . . . (α − n + 1) n
α
(1 + x) = 1 + αx +
x + ... +
x + o(x n)
2!
n!
n
x2 x3
n−1 x
ln(1 + x) = x −
+
+ . . . + (−1)
+ o(x n)
2
3
n
x
Note: ϕ(x) = o[ψ(x)] means that, as x → a, the function ϕ(x) has a higher order os smallness than the
ϕ(x)
=0
function ψ(x), i.e. lim
x→a ψ(x)
Note: a = e
ln a
x
. So a = e
x ln a
x 2(ln a)2
= 1 + x ln a +
+…
2!
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(1 + x) − (1 − x)
2x
e x − e −x
lim
=2
= lim
= lim
x→0 sin x
x→0
x
x→0 x
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x → x0 + 0 A = lim = f(x0 + 0)
ϵ>0
x→x0+0
δ(ϵ) > 0
0 < x − x0 < δ(ϵ)
| f(x) − A | < ϵ
The limit to the left of the function f(x0 − 0) as x → x0 − 0 is defined in a similar way.
Find the one-sided limits of the functions:
−2x + 3 x ≤ 1
{3x − 5
x>1
if x > 1, as x → 1
f(x) =
Solution: Let x ≤ 1. Then f(x) = − 2x + 3. Hence
f(1 − 0) = lim f(x) = 1 is the limit to the left.
x→1−0
If x > 1 then f(x) = 3x − 5. Hence f(1 + 0) = lim f(x) = − 2 is
the limit to the right
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x→1+0
Find the one-sided limits of the functions:
f(x) =
1 − cos 2x
x
as x → 0
2 sin x
2 sin2 x
=
=
f(x) =
x
x
x
π
π
But sin x = sin x if 0 < x < , and = − sin x if − < x < 0.
2
2
sin x
Hence f(0−) = lim f(x) = lim − 2
=− 2
x→0−0
x→0−
x
sin x
= 2
And f(0+) = lim f(x) = lim 2
x→0+
x
x→0+
1 − cos 2x
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f(x) = cos
π
as x → 0
(x)
Solution: We make use of Heine’s Definition of limits
We choose two sequences xn =
1
1
and xn′ =
2n
2n + 1
(n=1,2,…)
Then lim xn = lim xn′ = 0 and
xn→∞
xn′→∞
lim f(xn) = lim cos 2πn = 1
n→∞
n→∞
lim f(xn′) = lim cos(2n + 1)
n→∞
n→∞
π
=0
2
Hence, function f(x) has no limit to the right at point 0; taking into account that f(x) is an even
function, we conclude that it has not limit to the left either.




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y = f(x)
x0
x0 ∈ X
lim f(x) = f(x0)
x→x0
A function f(x) is continuous at the point x0 if and only if f(x0 − 0) = f(x0 + 0) = f(x0)
The function f(x) is continuous on the set X if it is continuous at every point of this set.
Discontinuity of First Kind
The point x0 is called a discontinuity of the first kind of the function f(x) if there exist the limits to the right and
to the left and there are finite. If f(x0 − 0) = f(x0 + 0) ≠ f(x0) then x0 is called a removable discontinuity. If
f(x0 − 0) ≠ f(x0 + 0) then x0 is called a jump discontinuity of the function f(x) at the point x0
Discontinuity of Second Kind
If at least one of the limits of f(x0 − 0) and f(x0 + 0) is non-existent and infinite, then the point x0 is called a
discontinuity of the second kind of the function f(x)
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f(x) =
2
(2x + 3) −∞ < x ≤ 1
6 − 5x
1<x<3
x−3
3≤x<∞
1
5
Find the points of discontinuity (if any).
Ashani Ray Classes (ashaniray.classes@gmail.com)
f(x) =
2
(2x + 3) −∞ < x ≤ 1
6 − 5x
1<x<3
x−3
3≤x<∞
1
5
Find the points of discontinuity (if any).
Solution: The domain of definition of the function is the entire number scale (−∞, ∞). In the open intervals
(−∞,1),(1,3) and (3,∞) the function is continuous. Therefore discontinuities are possible only at the points x=1,
x=3, at which the analytic representation of the function is changed.
1
f(1 − 0) = lim (2x 2 + 3) = 1,
x→1−0 5
f(1 + 0) = lim (6 − 5x) = 1.
x→1+0
The value of the function at the point x = 1 is determined by the first analytic representation, i.e. f(1) =
Since f(1 − 0) = f(1 + 0) = f(1), the function is continuous at the point x = 1
2+3
=1
5
Consider the point x = 3.
f(3 − 0) = lim (6 − 5x) = − 9 and f(3 + 0) = lim (x − 3) = 0 and f(3) = (3 − 3) = 0
x→3−0
x→3+0
We see that f(3 − 0) ≠ f(3 + 0) and the function had discontinuity of first kind (jump discontinuity) at x = 3
Jump of the function is f(3 + 0) − f(3 − 0) = 9
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Ashani Ray Classes (ashaniray.classes@gmail.com)
| 2x − 3 |
Given the function f(x) =
find the points of discontinuity (if any).
2x − 3
2x − 3 > 0
f(x) =
1
x>
for x >
−1 for x <
3
2
3
2
3
2
2x − 3 < 0
x<
3
2
3
3
+ 0 = 1 and f
−0 =−1
(2
)
(2
)
3
Therefore at x = the function has a finite discontinuity of the first kind.
2
3
3
+0 −f
− 0 = 1 − (−1) = 2
The jump of the function at this point is f
(2
)
(2
)
Hence f
x=
3
2
f(x) =
sin x
x
{1
x≠0
x=0
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f(x) =
sin x
x
{1
x≠0
x=0
Solution: The function is defined and continuous at all points x ≠ 0
sin x
sin x
= lim
= 1.
x→−0 x
x→+0 x
At the point x = 0 we have f(0) = 1; lim
Hence at this point the function is continuous as well, which means that it is continuous for all values of x.
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f(x) = sin
1
(x)
Ashani Ray Classes (ashaniray.classes@gmail.com)
f(x) = sin
1
(x)
Solution: The function is defined and continuous at all points x ≠ 0
There are no one-sided limits at the point x = 0 (See previous problem on f(x) = cos
Therefore at the point x = 0 the functions suffers a discontinuity of the second kind.
Ashani Ray Classes (ashaniray.classes@gmail.com)
π
)
(x)
f(x) =
x+2 x<2
{x 2 − 1 x ≥ 2
Ashani Ray Classes (ashaniray.classes@gmail.com)
x=2
f(x) =
x+2 x<2
{x 2 − 1 x ≥ 2
x=2
Solution: Find the one one-sided limits at the point x = 2
f(2 − 0) = lim (x + 2) = 4
x→2−0
f(2 + 0) = lim x 2 − 1 = 3
x→2+0
Hence the limits to the right and the left exist, are finite but do not coincide, therefore the function has a
discontinuity of the first kind at the point x = 2
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Continuity of a function
f(x) = sin x
Solution:
f(x) = sin x
Clearly the function f(x) is defines at every real number c and its value at c is sin c. We also know
that lim f(x) = lim sin x = sin c
x→c
x→c
Thus lim f(x) = f(c) and hence f(x) is continuous at every real number. This means that f(x) is a
x→c
continuous function.
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Continuity of a function
• f(x) = x
• f(x) = sin x
• f(x) = cos x
• ex
• ln x [Note: Domain of function ln x is (0, + ∞) ]
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f(x)
• f(x) ± g(x)
• f(x) ⋅ g(x)
f(x)
• g(x) [g(x0) ≠ 0]
g(x)
x = x0
Are also continuous at this point.
If the function u = ϕ(x) is continuous at the point x = x0 and the function y = f(u) is continuous at the
point u0 = ϕ(x0), then the composite function y = f [ϕ (x)] is continuous at the point x = x0
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2x 5 − 8x 2 + 11
f(x) = 4
x + 4x 3 + 8x 2 + 8x + 4
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2x 5 − 8x 2 + 11
f(x) = 4
x + 4x 3 + 8x 2 + 8x + 4
Solution: A function representing a ratio of two continuous functions (polynomials in this case) is discontinuous
only at points for which the denominator becomes zero.
But in our case x 4 + 4x 3 + 8x 2 + 8x + 4 = (x 2 + 2x + 2) and since x 2 + 2x + 2 = (x + 1)2 + 1 > 0 for any
x, the denominator never becomes zero. Hence the function f(x) is continuous throughout the entire number scale.
2
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3 sin3 x + cos2 x + 1
f(x) =
4 cos x − 2
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3 sin3 x + cos2 x + 1
f(x) =
4 cos x − 2
Solution: The function f(x) suffers discontinuities only at points for which the denominator equals zero, i.e. at
points which are the roots of the equation 4 cos x − 2 = 0 or cos x =
π
±
Whence x = xn = 2πn
, (n = 0, ± 1, ± 2,...)
3
Thus the function f(x) is continuous everywhere except at the points xn
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1
2
f(x) =
x 3 cos x + x 2 sin x
cos ( sin x )
1
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f(x) =
x 3 cos x + x 2 sin x
cos ( sin x )
1
Solution: The numerator of the function f(x) is continuous throughout the entire number scale.
As far as the denominator is concerned:
•
•
1
1
cos
it is continuous at points where the function u =
is continuous, since the function cos u is
( sin x )
sin x
continuous everywhere. Hence the denominator is continuous everywhere, except the the points x = kπ, where
k is an integer.
1
π
1
= (2p + 1) ,
= 0, i.e. the points at which
Besides, we must exclude the points at which cos
( sin x )
sin x
2
2
p ∈ ℤ, or sin x =
(2p + 1)π
Thus the function f(x) is continuous everywhere except at the points:
•
•
x = kπ, where k ∈ ℤ
x = (−1)n sin−1
2
+ nπ, where n, k ∈ ℤ
( (2p + 1)π )
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y=
1
1
where u =
u2 + u − 2
x−1
Solution: The function u = ϕ(x) =
1
suffers a discontinuity at the point x = 1.
x−1
1
suffers a discontinuity at the points where u 2 + u − 2 = 0, i.e. u1 = − 2
u2 + u − 2
1
and u2 = 1. Using these values of u, we find the corresponding values of x by solving the equations, −2 =
,
x−1
1
1
and 1 =
, from which we get x = and x = 2.
x−1
2
1
Hence, the composite function is discontinuous at three points: x1 = , x2 = 1, and x3 = 2.
2
The function y = f(u) =
To find the nature of discontinuity, we evaluate:
At x = 1: lim y = lim y = 0, therefore x = 1 is a removable discontinuity
•
•
•
x→1
y→∞
1
1
: lim y = lim y = ∞, therefore at x = there is discontinuity of second kind
u→−2
2 x→ 12
2
Atx = 2: lim y = lim y = ∞, therefore at x = 2 there is discontinuity of second kind
At x =
x→2
u→1
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x=1
If x ≠ 1, then u = f (f(x)) =
u = f (f(x))
v = f(x) =
1
1−
1
1−x
=
If x ≠ 0, x ≠ 1, then y = f (f (f (x))) =
1
1−x
x−1
. Then the point x = 0 is a discontinuity of the function
x
1
1−
x−1
x
= x, is continuous everywhere
Thus the points of discontinuity of this composite function are x = 0, x = 1, both of them being removable.
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f(x)
[a, b]
1. f(x) is bounded on [a, b]
2. f(x) has a minimum and maximum values on [a, b]
3. If m = min f(x), M = max f(x), then for any A satisfying the inequalities m ≤ A ≤ M there exists a
a≤x≤b
a≤x≤b
point x0 ∈ [a, b] for which f(x0) = A
In particular if f(a) ⋅ f(b) < 0, then we can find a point c where a < c < b, such that f(c) = 0
Intermediate value theorem
If f(x) is continuous on a closed interval [a, b] and c is any number between f(a) and f(b), inclusive, then
there is at least one number x in the interval [a, b] such that f(x) = c
Continuity of an Inverse function
If the function y = f(x) is defined, continuous, and strictly monotonic on the interval X, then there exists a
single valued inverse function x = ϕ(y) defined, continuous and also strictly monotonic in the range of the
function y = f(x)
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If −1 < x < 0, then −1 < x 2k+1 < 0 and 0 < x 2k < 1, where k ∈ ℕ.
Therefore [x r] = − 1 for odd values of r , and [x r] = 0 for even values of r, when −1 < x < 0
∴
2n+1
∑
r=1
r
[x ] =
n
∑
[x
2k+1
k=0
]+
n
∑
2k
[x ] =
k=1
n
∑
(−1) +
k=0
n
∑
(0) = − (n + 1), when −1 < x < 0
k=1
For x < 0, | x | = − x
2n+1
Therefore lim
x→0−
∑r=1 [x r] + (n + 1)
1 + [x] + | x | + 2x
−(n + 1) + (n + 1)
0
=0
= lim
x→0−
1 + 0 − x + 2x
x→0− x + 1
= lim
If −1 < x < 0, then −1 < x 2k+1 < 0 and 0 < x 2k < 1, where k ∈ ℕ.
Therefore [x r] = − 1 for odd values of r , and [x r] = 0 for even values of r, when −1 < x < 0
∴
2n+1
∑
r=1
[x r] =
n
∑
[x 2k+1] +
k=0
n
∑
[x 2k] =
k=1
n
∑
(−1) +
k=0
n
∑
(0) = − (n + 1), when −1 < x < 0
k=1
For x < 0, | x | = − x
2n+1
Therefore lim
x→0−
∑r=1 [x r] + (n + 1)
1 + [x] + | x | + 2x
−(n + 1) + (n + 1)
0
= lim = 0
x→0−
1 − 1 − x + 2x
x→0− x
= lim
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Problem: Evaluate lim
x→2
lim
2x + 23−x − 6
2−x − 21−x
2x = a
2x + 23−x − 6
x→2
= lim
a→4
= lim
a→4
= lim
a→4
= lim
a→4
x→2
a → 22
a→4
2−x − 21−x
a+
1
a
8
a
−6
−
2
a
a 2 − 6a + 8
a−2
(a − 2)(a − 4)
a−2
(a − 2) ( a + 2) ( a − 2)
( a − 2)
= lim (a − 2) ( a + 2) = (4 − 2) × (2 + 2) = 2 × 4 = 8
a→4
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Problem: If lim
x→−2−
ae
1
|x + 2|
2−e
−1
1
|x + 2|
x 4 − 16
, find the value of a
= lim + sin
x→−2
( x 5 + 32 )
x 4 − 16
lim sin
x→−2+
( x 5 + 32 )
= lim + sin
x→−2
(x + 4)(x + 2)(x − 2)
( (x + 2)(x 4 − 2x 3 + 4x 2 − 8x + 16) )
2
(x 2 + 4)(x − 2)
= lim + sin
x→−2
( (x 4 − 2x 3 + 4x 2 − 8x + 16) )
(8)(−4)
2
2
= sin
= sin −
= − sin
( (80) )
( 5)
(5)
Equating the limits obtained above
2
(5)
2
⟹ a = sin
(5)
−a = − sin
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Let | x + 2 | = y. Therefore as x → − 2−,
x + 2 → 0−, and | x + 2 | → 0+ and hence y → 0+
1
∴ lim −
x→−2
= lim+
y→0
= lim+
ae | x + 2 | − 1
1
2 − e |x + 2|
1
ae y − 1
1
2 − ey
− 1y
a−e
− 1y
2e − 1
a−0
=
=−a
2⋅0−1
y→0
x→∞ [
Problem: Evaluate lim
x→∞ [
lim
x→∞ [
= lim
= lim
x→∞
x+
x+
x+
x+
x+
x+
x+
x+
x+
x−
x ⋅
]
x−
x −x
x+
x+
x−
x
[
x+
x+
x+
x
]
[
x+
x+
x+
x
]
= lim
x
x→∞
x+
= lim
x→∞
1+
x+
x
1
x
x
=
]
]
[Dividing both numerator and denominator by
1+
x
1
1
=
1+1 2
+1
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x]
x+
x
x+
x+
x
2 x + 3 x + ... + n x
3
Problem: Evaluate lim
x→∞
2x − 3 +
3
2x − 3 + . . . +
2 x + 3 x + ... + n x
3
lim
2x − 3 +
x→∞
2+
= lim
x→∞
=
2
2
2−
=
n
3
x
+
3
n
2x − 3 + . . . +
3 x + ... + n x
3
n
x
3
+ 2x − 3 + . . . +
n
2x − 3
x
2
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n
2x − 3
n
2x − 3
Problem: Let f(x) = lim
n→∞
3
(π
1
tan−1 2x) + 5
2n
find the set of values of x for which f(x) = 0
f(x) = 0
3
tan−1 2x > 1
π
π
−1
tan 2x >
3
π
π
or tan−1 2x < −
tan−1 2x >
3
3
π
π
or
2x > tan
2x < − tan
3
3
3
3
x>
or x < −
2
2
3
x >
2
Therefore
⟹
⟹
⟹
⟹
⟹
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3
tan−1 2x
π
X
100x
99 sin x
+
[ x ])
x→0 ([ sin x ]
100x
99 sin x
= lim
+ lim
x→0 [ sin x ]
x→0 [
x ]
x
sin x
= lim 100 ⋅
+ lim 99 ⋅
x→0 [
sin x ] x→0 [
x ]
= 100 + 98
= 198
lim
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lim+ (1 − e x)
x→0 [
sin x
x ]
Let y = 1 − e x As, x → 0, we have y → 0−
sin x
As, x → 0, we have z → 1−
|x|
And yz → 0−
Let z =
sin x
Therefore lim (1 − e )
=−1
x→0+ [
x ]
x
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y 2 − 4y + 11 = y 2 − 4y + 4 + 7 = (y − 2)2 + 7
min (y 2 − 4y + 11) = 7
sin x
2
lim min (y − 4y + 11)
x→0 [
x ]
sin x
=6
= lim 7 ⋅
[
]
x→0
x
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{x} = x − [x]
Therefore lim {(1 + x) x }
2
x→0
= lim (1 + x) x − [(1 + x) x ]
)
x→0 (
2
2
= lim (1 + x) x − lim [(1 + x) x ]
2
x→0
2
x→0
= lim (1 + x) x − lim [(1 + x) x ]
x→0
2
= e − [e 2]
= e2 − 7
2
2
x→0
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1
ex
x
x
(2 ) − (3 )
lim
x→∞
xn
n
= lim
x→∞
3 (e
xn
ex
ex
xn
ex
⋅
xn
ex
xn
ex
ln
2
3
n
− 1)
⋅ ln
2
3
3
= lim
⋅ lim
x→∞ ( e x ) x→∞
(e
1
ex
xn
ex
ln
2 −3
= lim
x→∞
xn
2
3
⋅ ln
xn
ex
xn
ex
2
3
= lim
x→∞
a = e ln a
− 1)
⋅ ln
xn
ex
2
3
⋅ ln
3 (2 − 1)
xn
ex
xn
ex
xn
x
a x = e ln a = e x ln a
2
3
xn
Let y = .
ex
nx n−1
n(n − 1)x n−2
n!
xn
=0
=
lim
=
lim
=
lim
Now, lim
x→∞ e x
x→∞ e x
x→∞ e x
x→∞
ex
Therefore when x → ∞, we have y → 0
y
e
− 1)
2
1
2
1
(
y
⋅ lim x = ln ⋅ 1 ⋅ 1 ⋅ lim x = 0
Therefore above expression = ln ⋅ lim 3 ⋅ lim
3 y→0
y→0
y
x→∞ e
3
x→∞ e
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ax + 1 if x ≤ 3
{bx + 3 for x > 3
is continuous at x = 3
f(x) =
Solution:
f(3 − 0) = lim− (ax + 1) = 3a + 1
x→3
f(3 + 0) = lim+ (bx + 3) = 3b + 3
x→3
f(3) = 3a + 1
For function to be continuous
3a + 1 = 3b + 3
⟹ 3a = 3b + 2
2
⟹ a=b+
3
x
2
1
sin
x
1
is bounded between −1 and +1
x
We know for x ≠ 0,
1
−1 ≤ sin
≤ 1,
(x)
1
⟹ − x 2 ≤ x 2 sin
≤ x2
(x)
1
⟹ lim − x 2 ≤ lim x 2 sin
≤ lim x 2
( x ) x→0
x→0
x→0
sin
1
⟹ 0 ≤ lim x sin
≤0
(x)
x→0
1
Therefore lim x 2 sin = 0
x→0
x
Also, f(0) = 0.
2
Hence function is continuous
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x≠0
2
lim x − ax + b ≠ 0
x→1
x 2 − ax + b
lim
x→1
x−1
∞
Therefore lim x 2 − ax + b = 0, or 12 − a ⋅ 1 + b = 0 ⟹ a − b = 1
x→1
x 2 − ax + b
2x − a
Method 1: Using L’Hospital, lim
= lim 2x − a = 2 ⋅ 1 − a = 2 − a = 5
= lim
x→1
x→1
1
x→1
x−1
Therefore a = − 3, and b = − 4. Therefore a + b = − 7
x 2 − ax + b
= 5 Also, since lim x 2 − ax + b = 0, x 2 − ax + b = (x − 1)(x − α)
Method 2: lim
x→1
x−1
x→1
(x − 1)(x − α)
x 2 − ax + b
Therefore lim
= lim(x − α) = 1 − α = 5, making α = − 4
= lim
x→1
x−1
x→1
x→1
x−1
Therefore x 2 − ax + b = (x − 1)(x + 4) = x 2 + 3x − 4 = x 2 − ax + b
Comparing coefficients, a = − 3, b = − 4, making a + b = − 7
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(1) = 2
f(3) = f′(3) = 1
h(1)
Solution: f(1) = 2, Therefore g(2) = 1
f(3) = 1, Therefore g(1) = 3
dg df
dg
1
dg( f(x))
1
⋅
=1⟹
= df ⟹
=
x = g( f(x)) ⟹
df dx
df
df(x)
f′(x)
dx
Replace x by f −1, since f( f −1(x)) = x we have g (f (f −1(x))) = g (x) And
⟹
dg
1
=
dx
f′ (f −1(x))
h(x) = xf(x) + g(x)
h′(x) = f(x) + xf′(x) + g′(x) = f(x) + xf′(x) +
h′(1) = f(1) + f′(1) +















f′( f −1(x))
= f(x) + xf′(x) +
1
1
1
1
=2+2+
=4+
=4+ =5
f′(g(1))
f′(3)
f′(3)
1
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1
1
f′(g(x))
Problem: The number of real roots of the equation 1 + a1x + a2 x 2 + . . . + an x n = 0 where | x | <
| an | < 2, is
|x| <
1
3
f(x) = 1 + a1x + a2 x 2 + . . . + an x n
f(x) is min when − (a1x + a2 x 2 + . . . + an x n) is maximum
This will happen when for all ak x k has a negative value with | ak | = 2
Minimum value of f
2
2
2
1
−
= 1 − − 2 + ... = 1 −
( 3)
3 3
3
f′(x) = a1 + 2a2 x + 3a3x 2 + . . . + nan x n−1

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1
1−
1
3
1
and
3
lim
x→0
2. lim
3
sin 3x
5
5x
2x + 23−x − 6
2−x − 21−x
x→2
3.
x→∞ [
lim
x+
5.
6.
7.
3
8. lim
x→0
9.
x→4
[Ans: 8]
x+
1
ex
n
1+x−
x
x+2
lim
x→∞ ( x + 1 )
3
x−
1
x [Ans: ]
2
]
1
ex
1−x
x+3
3−
5+x
1−
5−x
1
3
[Ans: − ]
G(x) − G(1)
2. If G(x) = − 25 − x , evaluate lim
x→1
x−1
1
xa − aa
3. lim
[Ans:
]
x→a a x − a a
ln a
2
x
x
(2 ) − (3 )
[Ans: 0]
lim
n
x→∞
x
15x − 5x − 3x + 1
lim
[Ans: ln 3 ln 5]
x→0
sin2 x
1
3x+1 − 3
lim
[Ans: ln 3]
3
x→0
9x
1
ln x − ln 5
lim
[Ans: ]
5
x→5
x−5
n
4.
1. lim
2
3
[Ans: ]
[Ans: e]
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4. lim
x→0
1
2 sin( x )
x e
[Ans: 0]
2 x + 3 x + ... + n x
3
5.
lim
x→∞
2x − 3 +
[Ans:
2]
3
2x − 3 + . . . +
n
n
2x − 3
sin 3x
x→0 5x
sin 3x
sin 3x 3x
= lim
⋅
Solution: lim
x→0 5x
x→0 3x
5x
sin 3x
3x
= lim
⋅ lim
[Since as x → 0, 3x → 0]
3x→0 3x
x→0 5x
3 3
=1⋅ =
5 5
lim
3x+1 − 3
Problem: lim
x→0
9x
x
x ln 3
x ln 3
3 (3 x − 1 )
− 1)
− 1) ln 3
3x+1 − 3
(3 − 1)
(e
(e
= lim
= lim
= lim
= lim
⋅
Solution: lim
x→0
9x
x→0
9x
x→0
3x
x→0
3x
x→0
3x
ln 3
x ln 3
− 1)
1
(e
Let x ln 3 = y When x → 0, we have x ln 3 → 0, i.e. y → 0. Hence above expression
= ln 3 lim
3
x→0
x ln 3
y
1
( e − 1) 1
is equal to: = ln 3 lim
= ln 3
3
y→0
y
3
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lim
x→4
Solution: lim
x→4
x→4 ( 1
= lim
= lim
x→4
3−
5+x
1−
5−x
3−
5+x
1−
5−x
3−
5+x
−
5−x
⋅
1+
5−x
3+
=−
Solution: lim
x→5
3−
5+x
1−
5−x
⋅
)
3+
5−x
5+x
1+
5+x
ln x − ln 5
Problem: lim
x→5
x−5
x→4
x→4
3+
9 − (5 + x) 1 +
⋅
1 − (5 − x) 3 +
= − lim
= lim
5−x
5+x
1+
= lim
x→4
⋅
3+
5+x
3+
5+x
⋅
1+
5−x
1+
5−x
5−x
5+x
4−x 1+
⋅
x−4 3+
5−x
5+x
= − lim
x→4
1+1
1
=−
3+3
3
ln x − ln 5
ln(z + 5) − ln 5
Let z = x − 5 = lim
= lim
x−5
z→0
z
z→0
1 ln ( 5 + 1) 1
=
= lim ⋅
z
z→0 5
5
z
5
x−4 1+
⋅
x−4 3+
Ashani Ray Classes (ashaniray.classes@gmail.com)
ln (
z+5
5 )
z
5−x
5+x
3
lim
1+x−
3
1+x−
( 1+x−
3
= lim
x→0
= lim
x→0
= lim
x→0
3
x
x→0
3
1−x
x
x→0
= lim
1−x
x
x→0
Solution: lim
3
1 − x) ( 1 + x) +
⋅
2
3
( 1 + x) +
3
2
3
1 + x 1 + x + ( 1 + x)
3
1 + x 1 + x + ( 1 + x)
3
3
(1 + x) − (1 − x)
x ( 1 + x) +
[
2
3
x ( 1 + x) +
[
2
3
[(
3
1 + x 1 + x + ( 1 + x)
3
1 + x 1 + x + ( 1 + x)
3
3
1 + x) +
2
3
1 + x 1 + x + ( 1 + x)
3
Ashani Ray Classes (ashaniray.classes@gmail.com)
3
]
]
2
3
2
2
3
2
3
2x
3
2
3
]
2
=
( 1) +
3
2
2
3
1 1 + ( 1)
3
3
2
=
2
3
3
lim
3
1−x
3
1−x
x
x→0
3
Solution: lim
x→0
1+x−
1
1+x−
x
3
lim
1+x−1−
x→0
1
(1 + x) 3 − 1
(1 − x) 3 − 1
= lim
− lim
x→0
x
x→0
x
3
1−x+1
x
Let x + 1 = y. Therefore x = y − 1 and as x → 0 we have y → 1
Let 1 − x = z. Therefore x = 1 − z and as x → 0 we have z → 1
1
3
1
1
3
1
(z) 3 − 1
(y) − 1
= lim
− lim
y→1 y − 1
z→1 1 − z
(z) 3 − 1 1 1 2
(y) − 1
= lim
+ lim
= + =
y→1 y − 1
z→1 z − 1
3 3 3
Ashani Ray Classes (ashaniray.classes@gmail.com)
G(x) = −
25 − x 2
lim
x→1
G(x) − G(1)
x−1
− 25 − x 2 + 25 − 1
G(x) − G(1)
= lim
Solution: lim
x→1
x−1
x→1
x−1
= lim
x→1
= lim
x→1
24 −
25 − x 2
24 −
25 − x 2
x−1
⋅
x−1
24 − 25 + x 2
= lim
⋅
x→1
x−1
x2 − 1
= lim
⋅
x→1 x − 1
24 +
25 − x 2
24 +
25 − x 2
1
24 +
25 − x 2
1
24 +
25 − x 2
= lim (x + 1) ⋅
x→1
Ashani Ray Classes (ashaniray.classes@gmail.com)
1
24 +
25 − x 2
=
2
24 +
24
=
1
24
xa − aa
lim x
x→a a − a a
xa − aa
xa − aa
x−a
Solution: lim
=
lim
⋅
x→a a x − a a
x→a x − a
a a (a x−a − 1)
xa − aa
x−a
= lim
⋅
x→a x − a
a a (e (x−a)ln a − 1)
xa − aa
(x − a)ln a
1
= lim
⋅
⋅ a
x→a x − a
(e (x−a)ln a − 1) a ln a
xa − aa
z
1
= lim
⋅ lim z
⋅ a
[Substituting z = x − a]
x→a x − a
z→0 (e − 1) a ln a
xa − aa
= lim
⋅ lim
x→a x − a
z→0
= (a ⋅ a
1
ez − 1
z
⋅
1
a a ln a
1
1
=
)⋅1⋅ a
a ln a ln a
a−1
Ashani Ray Classes (ashaniray.classes@gmail.com)
x+2
lim
x→∞ ( x + 1 )
x+3
x+2
Solution: lim
x→∞ ( x + 1 )
x+3
1
= lim 1 +
x→∞ (
x + 1)
x+3
Let x + 1 = y Hence when x → ∞ we have y → ∞.
1
lim 1 +
x→∞ (
x + 1)
x+3
1
= lim 1 +
y→∞ (
y)
y+2
1
= lim 1 +
y→∞ (
y)
y+2
1
1
= lim 1 +
⋅ lim 1 +
= e ⋅ (1 + 0)2 = e
y→∞ (
y ) y→∞ (
y)
y
2
Alternate:
1
1+
y→∞ (
y)
lim
=e
y+2
lim y→∞ 1y ⋅(y + 2)
1
1+
y→∞ [(
y) ]
= lim
=
lim y→∞(1 + 2y )
e
y
1
y ⋅(y
+ 2)
1
1+
y→∞ [(
y) ]
= lim
= e 1+0 = e 1 = e
Ashani Ray Classes (ashaniray.classes@gmail.com)
1
y lim y→∞ y ⋅(y + 2)
x3 + x − 2
lim 3
x→1 x − x 2 − x + 1
x3 + x − 2
x 3 − x 2 + x 2 − x + 2x − 2
(x − 1)(x 2 + x + 2)
Solution: lim
= lim
= lim
3
2
2
x→1 x − x − x + 1
x→1
(x − 1)(x − 1)
x→1 (x − 1)(x + 1)(x − 1)
x 2 + x + 2)
= lim
=∞
x→1 (x + 1)(x − 1)
x3 + x − 2
Problem [Berman Ch3/277]: lim
x→1 x 3 − x 2 − x + 1
x2 + x − 2
1 + x + x2 − 3
1
3
= lim
−
= lim
Solution: lim
3
3
(
)
x→1
1−x 1−x
x→1 (
1−x
) x→1 (1 − x)(1 + x + x 2)
(x − 1)(x + 2)
−(x + 2)
x 2 − x + 2x − 2
= lim
= lim
= lim
=−1
2
2
2
x→1 (1 − x)(1 + x + x )
x→1 (1 − x)(1 + x + x )
x→1 1 + x + x
Ashani Ray Classes (ashaniray.classes@gmail.com)
xm − 1
lim n
x→1 x − 1
xm − 1
xm − 1
x−1
xm − 1
= lim
× n
= lim
× lim
Solution: lim
n
x→1 x − 1
x→1 x − 1
x − 1 x→1 x − 1
x→1
1 m
xm − 1
1
=
m
×
= lim
×
=
n−1
x
n
n
x→1 x − 1
limx→1 x − 1
(x − 1)(1 + x + x 2 + . . . + x m−1)
xm − 1
= lim
Alternate: lim
x→1 x n − 1
x→1 (x − 1)(1 + x + x 2 + . . . + x n−1)
1 + x + x 2 + . . . + x m−1 m
=
= lim
2
n−1
n
x→1 1 + x + x + . . . + x
Ashani Ray Classes (ashaniray.classes@gmail.com)
1
xn − 1
x−1
sin x 2
sin x 2
x
1
sin x 2
x
= lim 2 ⋅
= lim
⋅
⋅ cos x
lim f(x) = lim
x→0 2x tan x
x→0 x
2 tan x 2 x→0 x 2
sin x
x→0
1
sin x 2
1
1
1
= lim 2 ⋅ lim sin x ⋅ lim cos x = 1 ⋅ 1 ⋅ 1 =
2 x→0 x
x→0
x→0
2
2
x
1
Since function is continuous, f(0) =
2
1
Hence a =
2
Ashani Ray Classes (ashaniray.classes@gmail.com)
Problem[2019 Main, 8 April II]: Let f : ℝ → ℝ de a differentiable function satisfying
1
x
1 + f (3 + x) − f (3)
is equal to : (a) e (b) e −1 (c) e 2 (d) 1
x→0 ( 1 + f (2 − x) − f (2) )
f′(3) + f′(2) = 0. Then lim
f(x)
lim f(x + h) = f(x)
h→0
Hence it is of form 1∞
1
x
+ x) − f(3)
1
1 + f (3 + x) − f (3)
limx→0( 11 ++ f(3
f(2 − x) − f(2) − 1)⋅ x
lim
=e
(
)
x→0
1 + f (2 − x) − f (2)
1 + f (3 + x) − f (3)
1
1
[ f(3 + x) − f(3)] + [ f(2) − f(2 − x)]
⋅
= lim
− 1 ⋅ = lim
) x x→0 (
x→0 ( 1 + f (2 − x) − f (2)
1 + f (2 − x) − f (2)
) x
f(3 + x) − f(3)
= lim
x→0
x
+
f(2) − f(2 − x)
f′(3) + f′(2)
=0
= lim
x→0
1
x
1 + f (2 − x) − f (2)
1
x
1 + f (3 + x) − f (3)
= e0 = 1
x→0 ( 1 + f (2 − x) − f (2) )
Hence lim




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Ashani Ray Classes (ashaniray.classes@gmail.com)
Problem[2019 Main, 11 Jan I]: Let [x] denote the greatest integer function less than or equal to
x. Then lim
tan (π sin2 x) + ( x − sin (x[x]))
x2
x→0
tan (π sin x)
lim
= lim
x→0
x2
x→0
2
= lim
x→0
sin (π ⋅ x 2)
πx 2
⋅
π
=π
2
cos πx
tan (π ⋅ x 2 ⋅
2
equals/does not exist:
sin x
x
⋅
x2
= lim
x→0
( x − sin (x[x]))
(x − sin (0))
lim
= lim
=1
2
2
x→0+
x
x→0+
x
2
sin x
x )
tan (π ⋅ x 2)
x2
2
( x − sin (x[x]))
(−x − sin (−x))
(−x + sin x)
lim
= lim
= lim
x→0−
x2
x→0+
x2
x→0+
x2
x 2 + sin2 x − 2x sin x
= lim
=0
2
x→0+
x
2
2
2
Ashani Ray Classes (ashaniray.classes@gmail.com)
Problem: Discuss the continuity of the function f defined by f(x) = | x − 1 | + | x − 2 | at
x = 1 and x = 2
1
lim x − a sin
=0
(
)
x→a−
x−a
lim x − a sin
x→a+
1
=0
(x − a)
Let x − a = h . As x → a− we have h → 0−
1
1
lim h sin
= lim h sin
=0
(
)
(
)
h→0−
h
h→0−
h
Problem: Evaluate lim
x→0
1
2 sin( x )
x e
1
−1 ≤ sin
≤1
(x)
⟹ e
−1
≤e
2 −1
⟹ x e
sin( 1x )
≤
≤ e1
1
2 sin( x )
x e
2 −1
1
2 sin( x )
x e
⟹ lim x e
≤ lim
⟹ 0 ≤ lim
1
2 sin( x )
x e
x→0
x→0
Hence lim
x→0
x→0
≤ x 2e 1
≤ lim x 2e 1
x→0
≤0
1
2 sin( x )
is 0
x e
Ashani Ray Classes (ashaniray.classes@gmail.com)
6
x=y
⟹ x = y6 ⟹
lim
x→64
x = y 3 and ⟹
3
x = y 2 and as x → 64 we have y → 2
x−8
3
y3 − 8
(y − 2)(y 2 + 2y + 4)
y 2 + 2y + 4
= lim 2
= lim
= lim
y→2
y
−
4
y→2
(y
−
2)(y
+
2)
y→2
y+2
x−4
22 + 2 ⋅ 2 + 4
=3
=
2+2
2x + x
Problem: lim
x→∞ 3x
x
2x + x
2
x
lim
= lim
+ lim x = 0
x
(
)
x→∞ 3
x→∞
3
x→∞ 3
Ashani Ray Classes (ashaniray.classes@gmail.com)
lim
x→0
Solution: lim
x→∞
= lim
x→∞
(x + 1)10 − (x − 1)8
10
(x − 1) − (x + 1)
8
=−1
(x + 1)10 − (x − 1)8
10
(x − 1) − (x + 1)
8
= lim
x→∞
(x + 1)10 − (x − 1)8
(x − 1)10 − (x + 1)8
(x 10 + 10x 9 + . . . + 1) − (x − 1)8
(x − 1)10 − (x + 1)8
x 2 − 16
Problem: lim
x→4 x 2 − 7x + 12
(x − 4) (x + 4)
(x + 4)
x 2 − 16
=8
=
lim
=
lim
Solution: lim
2
x→4 (x − 4) (x − 3)
x→4 (x − 3)
x→4 x − 7x + 12
Ashani Ray Classes (ashaniray.classes@gmail.com)
lim
1
4
(x + h) − x 4
1
h
h→0
1
4
Solution: Let (x + h) = a and x 4 = b
1
⟹ x + h = a 4 and x = b 4
⟹ h = a4 − b4
And as h → 0 we have a 4 − b 4 → 0 or b 4 → a 4 or b → a
1
4
a−b
(x + h) − x
lim
= lim 4
a→b a − b 4
h→0
h
a−b
= lim
a→b (a − b)(a + b)(a 2 + b 2)
1
4
1
1
1
1
1 −3
= lim
=
= 3 = 3 = x 4
2
2
2
2
a→b (a + b)(a + b )
(b + b)(b + b ) 4b
4
4x 4
Ashani Ray Classes (ashaniray.classes@gmail.com)
2x − 1
lim x
x→0 4 − 1
ex − 1
= 1 and 2x = e x ln 2
Solution: We know lim
x→0
x
e x ln 2 − 1
e x ln 2 − 1
2x ln 2
x ln 2
2x − 1
2x − 1
= lim 2x ln 2
= lim
⋅ 2x ln 2
⋅
lim x
= lim 2x
x→0 e
− 1 x→0 x ln 2
e
− 1 2x ln 2
x→0 4 − 1
x→0 2 − 1
Ashani Ray Classes (ashaniray.classes@gmail.com)
2x 2 − 3cx + x + c
2x 2 − 3cx + x + c
lim
= lim
2
x→2
x −4
x→2 (x + 2) (x − 2)
For limit to exist, x − 2 must be a factor of 2x 2 − 3cx + x + c , i.e
2x 2 − 3cx + x + c = 0 when x = 2
⟹ 2 ⋅ 22 − 3c ⋅ 2 + 2 + c = 0
⟹ 10 = 5c
⟹ c=2
(x − 2) (2x − 1) 3
2x 2 − 3cx + x + c
2x 2 − 5x + 2
= lim
= lim
= lim
=
x→2 (x + 2) (x − 2)
x→2 (x + 2) (x − 2)
x→2 (x + 2) (x − 2)
4
Ashani Ray Classes (ashaniray.classes@gmail.com)
2x
lim
x→0 sin x + 2x
2x
= lim
Solution: lim
x→0 sin x + 2x
x→0
2
sin x
x
Ashani Ray Classes (ashaniray.classes@gmail.com)
+2
=
2
2
=
1+2 3
ax + b
y=
x+c
ax + b
y=
x−2
As x → ∞, y → 2
x→−c
x (a + bx )
y→∞
2+c=0 ⟹ c=−2
b
a
+
(
x)
ax + b
=a
= lim
= lim
lim
2
2
x→∞
x→∞
x→∞ x − 2
x (1 − x )
(1 − x )
Hence a = 2
2x + b
2x + 4
. It passes through (-2, 0) ⟹ b = 4 ⟹ y =
y=
x−2
x−2
Ashani Ray Classes (ashaniray.classes@gmail.com)
(2, 4, -2)
R(0,r)
x2 − 0
=
=x
x−0
P (x, x 2)
x x2
,
Mid point of OP: Q
(2 2 )
O
1
Slope of ⊥ bisector of OP = −
x
⟹
x2
r− 2
x
0− 2
1
=−
x
x2 1
⟹ r=
+
2
2
Ashani Ray Classes (ashaniray.classes@gmail.com)
x2 1
1
lim r = lim
+
=
x→0
x→0 ( 2
2) 2
P (x,0)
Ashani Ray Classes (ashaniray.classes@gmail.com)
f′′ > 0




When reaching/around a maxima f′′ < 0
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