1210-004 CALCULUS I
Review sheet for Midterm 1
January 24, 2014
Contents
1 Limits
1
2 Continuity
3
3 Derivatives
5
4 Examples
6
1
Limits
• Definition of limit: we say that limx→c f (x) = L if f (x) is near L when x is near c (but
different from c).
• One-sided limits:
1. We say that limx→c+ f (x) = L if f (x) is near L when x is near c and x > c.
2. We say that limx→c− f (x) = L if f (x) is near L when x is near c and x < c.
• Note that limx→c f (x) = L if and only if limx→c+ f (x) = limx→c− f (x) = L.
• Rules to compute limits: let k be a constant, n be a positive integer, c be a number,
and f and g be functions that have limits at c.
1. limx→c k = k.
2. limx→c x = c.
3. limx→c k · f (x) = k limx→c f (x).
4. limx→c [f (x) + g(x)] = limx→c f (x) + limx→c g(x).
5. limx→c [f (x) − g(x)] = limx→c f (x) − limx→c g(x).
6. limx→c [f (x) · g(x)] = limx→c f (x) · limx→c g(x).
7. limx→c
f (x)
g(x)
=
limx→c f (x)
limx→c g(x)
provided that limx→c g(x) 6= 0.
8. limx→c [f (x)]n = [limx→c f (x)]n .
1
9. limx→c
p
p
n
f (x) = n limx→c f (x) provided that limx→c f (x) ≥ 0 when n is even.
• Squeeze theorem: If f (x) ≤ g(x) ≤ h(x) and limx→c f (x) = limx→c h(x) = L, then
limx→c f (x) = L.
• Zero times bounded (See example 4.6 below): If limx→c f (x) = 0 and g(x) is
bounded near c, then limx→c f (x)g(x) = 0.
For example, limx→0 x sin x1 = 0 even though limx→0 sin x1 does not exist.
• Limits of trigonometric functions (See example 4.5 below):
1. limx→c sin x = sin c
2. limx→c cos x = cos c
3. limx→c tan x = tan c provided that cos c 6= 0.
4. limx→c cot x = cot c provided that sin c 6= 0.
5. limx→c sec x = sec c provided that cos c 6= 0.
6. limx→c csc x = csc c provided that sin c 6= 0.
sin x
x =
cos x−1
limx→0 x
7. limx→0
1.
8.
= 0.
• Limits at infinity (See example 4.6 below)
1. If k is a positive integer, limx→∞ xk = ∞.
2. If k is a positive integer, limx→∞
1
xk
= 0.
3. If f (x) = p(x)
q(x) is a rational function (quotient of polynomials), the behavior at infinity
is given by the leading monomials of p and q. To compute the limit, just divide
the numerator and the denominator by the highest power of x appearing in the
denominator.
p(x)
q(x)
p(x)
then limx→∞ q(x)
then limx→∞ p(x)
q(x)
(i) If deg p > deg q, then limx→∞
=∞
(ii) If deg p < deg q,
=0
(iii) If deg p = deg q,
= C where C is the quotient of the coefficients
of the higher-degree monomials of p and q.
• Infinite limits: these occur when limx→c f (x) = ±∞ (vertical asymptotes). One has to
determine the sign of ∞.
1. If f (x) = x1 , then limx→0+
2. If f (x) =
1
,
(x−1)2
1
x
= +∞ and limx→0−
then limx→1+
1
(x−1)2
2
1
x
= −∞.
= +∞ and limx→1−
1
(x−1)2
= +∞
2
Continuity
• A function f is continuous at x = c if limx→c f (x) = f (c). This means that 3 things
must happen:
1. f must be defined at c.
2. The limit limx→c f (x) must exist.
3. Both f (c) and limx→c f (x) must coincide.
• Continuity and operations:
1. If f is continuous at c and k is a constant, then k · f is continuous at c.
2. If f is continuous at c, then f n is continuous at c.
√
3. If f is continuous at c, then n f is continuous at c (provided that f (c) ≥ 0 when n is
even).
4. If f and g are continuous at c, then f + g is continuous at c.
5. If f and g are continuous at c, then f − g is continuous at c.
6. If f and g are continuous at c, then f · g is continuous at c.
7. If f and g are continuous at c, then
f
g
is continuous at c provided that g(c) 6= 0.
8. If g is continuous at c and f is continuous at g(c), then f ◦ g is continuous at c.
• A function is discontinuous at a point x = c if any of the above conditions fails to hold.
(See examples 4.1 and 4.2 below)
1. Example of function f not defined at c = 1 such that limx→1 f (x) does not exist:
f (x) =
1
x−1
(this is a vertical discontinuity).
2. Example of function f not defined at c = 1 such that limx→1 f (x) exists:
f (x) =
x2 − 3x + 2
x−1
(this is a removable discontinuity).
3. Example of function f defined at x = 1 such that limx→1 f (x) does not exist:
f (x) = [[x]]
(this is a jump discontinuity).
• Functions defined piecewise (See example 4.1 below)
g(x), x < c
f (x) =
h(x), x ≥ c
In order to study the continuity of functions which are defined piecewise like f (x) one has
to:
3
1. Study the continuity of each piece within its interval of definition (in the example
above, study the continuity of g(x) over (−∞, c) and study the continuity of h(x)
over (c, ∞).
2. Study the continuity at the endpoints of the intervals of definition. In the example
above, determine whether or not
lim f (x) = lim f (x)
x→c−
x→c+
To do this, note that when x → c− , f is given by f (x) = g(x), whereas when x → c+ ,
then f is given by f (x) = h(x).
• Intermediate value theorem (See example 4.3 below): if f is a continuous function which is defined and is continuous over an interval [a, b], then for any number M ∈
[f (a), f (b)] there exists some c ∈ [a, b] such that f (c) = M . In particular, if f (a) and f (b)
have opposite signs then M = 0 lies in between, so there exists some c ∈ [a, b] such that
f (c) = 0.
4
3
Derivatives
• The derivative of a function f (s) is another function f 0 (x) which is given at every x by
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
If this limit exists at x we say that the function f is differentiable at x.
• If the limit does not exist at some point x = c, we say that the function is not differentiable
at that point. For example, the function f (x) = |x| is not differentiable at x = 0 because
the above limit does not exist (the left-hand and right-hand limits do not coincide)
f (0 + h) − f (0)
|h| − |0|
|h|
= lim
= lim
h→0
h→0
h→0 h
h
h
lim
so limh→0−
|h|
h
= −1 but limh→0+
|h|
h
= 1.
• Geometrically, for every c, f 0 (c) is the slope of the tangent line to the curve y = f (x)
at the point x = c (See example 4.4 below).
• If f is differentiable at c, then f is continuous at c. The converse is not true, as the
absolute value function shows (it is continuous everywhere, but it is not differentiable at
0.
• Differentiation rules: computing derivatives using limits can be tedious, so it is good
to keep in mind a few simple rules:
1. Constant function: If f (x) = C, then f 0 (x) = 0.
2. Identity function: If f (x) = x, then f 0 (x) = 1.
3. Powers: If f (x) = xn with n > 0, then f 0 (x) = nxn−1 .
4. Constant multiples: If C is constant, then (C · f )0 (x) = C · f 0 (x).
5. Sums: If f and g are differentiable at x, then (f + g)0 (x) = f 0 (x) + g 0 (x)
6. Differences: If f and g are differentiable at x, then (f − g)0 (x) = f 0 (x) − g 0 (x)
7. Products: If f and g are differentiable at x, then (f · g)0 (x) = f 0 (x)g(x) + f (x)g 0 (x)
0
0
(x)g 0 (x)
8. Quotients: If f and g are differentiable at x, then fg (x) = f (x)g(x)−f
g 2 (x)
• Derivatives of trigonometric functions:
1. Dx (sin x) = cos x
2. Dx (cos x) = − sin x
3. Dx (tan x) = sec2 x (follows from (1) and (2), using the quotient rule).
4. Dx (cot x) = − csc2 x (follows from (1) and (2), using the quotient rule).
5. Dx (sec x) = sec x tan x (follows from (2), using the quotient rule).
6. Dx (csc x) = − csc x cot x (follows from (1), using the quotient rule).
5
4
Examples
These examples are taken from the 3 quizzes and illustrate some of the concepts above:
Example 4.1 (Continuity of a function defined piecewise). Consider the function
 x2 +x
x<0
 x+1 ,
2
f (x) =
x ,
0≤x≤1

3 − x,
x>2
(i) At which points is the function discontinuous?
(ii) For every point of discontinuity: if the discontinuity is removable, compute the limit at the
point; if the function has a vertical asymptote or a jump, compute the one-sided limits.
Solution: The function is defined piecewise, so it will have a discontinuity at any point
where any of the pieces has a discontinuity, and possibly at the points x = 0, 1 if the one-sided
limits do not coincide.
2
+x
When x < 0, the function is given by f (x) = xx+1
. It is continuous everywhere except at
x = −1, where the denominator vanishes. At this point the limit is given by
x2 + x
x(x + 1)
= lim
= lim x = −1
x→−1 x + 1
x→−1 x + 1
x→−1
lim
Since the limit at x = −1 exists, the discontinuity is removable.
The functions x2 and 3 − x are both continuous everywhere, f is continuous on the intervals
[0, 1] and [3, +∞].
At the points x = 0 and x = 1 where the definition of the function changes, the function will
be continuous only if both one-sided limits exist and coincide. At x = 0 we have
0
x2 + x
= = 0,
x→0 x + 1
1
lim f (x) = lim
x→0−
lim f (x) = lim x2 = 02 = 0
x→0+
x→0
so the function is continuous at x = 0. At x = 1 we have
lim f (x) = lim x2 = 12 = 1,
x→1−
lim f (x) = lim 3 − x = 2
x→1
x→1+
x→1
Since both limits differ, the function has a jump discontinuity at x = 1.
In summary: the function has a removable singularity at x = −1 and a jump discontinuity
at x = 1.
√
Example 4.2 (Study the continuity of a function). Consider the function f (t) =
t−1
(t−1)(t+1)(t−2) .
(i) What is the domain of f (i.e. where is f defined?)
(ii) At which points is the function discontinuous? Are the discontinuities removable / jumps
/ vertical?
6
(iii) For every removable discontinuity, find the limit.
(iv) For every jump or vertical discontinuity, find the corresponding one-sided limits.
Solution:
(i) The domain of the function f is all of [0, ∞], except for the
√ points t = 1 and t = 2 where
the denominator is 0. If t < 0, f (t) is not defined because t is not defined.
(ii) The function is discontinuous at the points t = 1 and t = 2 because the denominator is
zero. The denominator is also zero at t = −1, but this point is not in the domain of f in
the first place.
√
Evaluating at t = 2, wee see that f (2) =
2−1
0 ,
so there is a vertical discontinuity at t = 2.
Evaluating at t = 1, wee see that f (1) = 00 , so wee need to compute the limit. If the limit
exists, the discontinuity will be removable.
(iii) We evaluate the limit as t → 1:
√
√
t−1
t−1
√
= lim √
lim f (t) = lim
t→1 ( t + 1)( t − 1)(t + 1)(t − 2)
t→1
t→1 (t − 1)(t + 1)(t − 2)
1
1
= lim √
= √
t→1 ( t + 1)(t + 1)(t − 2)
( 1 + 1)(1 + 1)(1 − 2)
1
= −
4
(iv) We compute the left-hand and right-hand limits. As t → 2+ , every factor in the definition
of f (t) is positive, so limt→2+ f (t) = +∞.
As t → 2− , every factor in the definition of f (t) is positive except for t−2 which is negative,
so limt→2− f (t) = −∞
Example 4.3 (Application of the Intermediate Value theorem). Show that the equation (cos t)t3 +
6 sin5 t − 3 = 0 has a solution between t = 0 and t = 2π.
Solution: The intermediate value theorem states that if f is a continuous function which
is defined and is continuous over an interval [a, b], then for any number M ∈ [f (a), f (b)] there
exists some c ∈ [a, b] such that f (c) = M . In particular, if f (a) and f (b) have opposite signs
then M = 0 lies in between, so there exists some c ∈ [a, b] such that f (c) = 0.
Consider the function f (t) = (cos t)t3 + 6 sin5 t − 3, so that a solution of the equation
(cos t)t3 + 6 sin5 t − 3 over [0, 2π] is a number c ∈ [0, 2π] such that f (c) = 0. The function f is
continuous everywhere, so in particular it is continuous over [0, 2π]. Moreover, we have that
f (0) = (cos 0) · 03 + 6 sin5 0 − 3 = −3 < 0
f (2π) = (cos 2π) · (2π)3 + 6 · sin5 (2π) − 3 = 8π 3 − 3 > 0
Since f (0) < 0 < f (2π) and f is continuous over [0, 2π], the intermediate value theorem
guarantees that there exists c ∈ [0, 2π] such that f (c) = 0
7
Example 4.4 (Slope of the tangent line to a curve). Find the slope of the tangent line to the
1
curve y = x−1
at the point (0, −1).
Recall that if a curve is given by an equation y = f (x), the slope of the tangent line at a
point (c, f (c)) is given by the limit
f (c + h) − f (c)
h→0
h
lim
Solution: The slope of the tangent line at the point (0, −1) is given by the limit
f (x) − f (0)
lim
= lim
x→0
x→0
x−0
1
x−1
1
1
− 0−1
x
1
x−1 + 1
= lim
= lim
= lim
= −1
x→0 x − 0
x→0 x(x − 1)
x→0 x − 1
x−0
The equation of the tangent line will be given by y = −x − 1
Example 4.5 (Limits of trigonometric functions). Find limt→0
[tan(3t)]2
.
2t
Solution:
[tan(3t)]2
lim
t→0
2t
=
=
1
sin(3t) sin(3t)
1
lim lim
·
=
2
2 t→0 t→0
t
cos (3t)
2
3
0
·1· =0
2
1
√2x+1
x3 +2
Example 4.6 (limits at infinity). Find limx→∞
sin(3t)
sin(3t)
3 lim
· lim
t→0
t→0 cos2 (3t)
3t
and limx→∞
√2x+1
x3 +2
sin(x4 )
grow to
Solution: Both the numerator and the denominator of the rational function √2x+1
x3 +2
∞ as x does. The largest exponent in the numerator is 1, whereas the largest exponent in the
denominator is 23 , so the intuition tells us that the limit will be zero. To see this we divide both
the numerator
and the denominator by the top degree monomial appearing in the denominator,
√
namely x2 , so that
2x + 1
lim √
= lim
x→∞
x3 + 2 x→∞
√1 · (2x
x3
√
√1
x3
3
x
+ 1)
√2
x
+
√1
x3
= lim q
x→∞
+2
1+
2
x3
=
0
=0
1
In order to compute limx→∞ √2x+1
sin(x4 ), we observe that sin(x4 ) is a bounded function, so
x3 +2
even though limx→∞ sin(x4 ) does not exist, by the squeeze theorem
2x + 1
2x + 1
4
lim √
sin(x ) = lim √
· lim sin(x4 ) = 0
x→∞
x→∞
x3 + 2
x3 + 2 | x→∞ {z
}
|
{z
}
→0
8
bounded