Calculate the limit of a function: Slightly More Complicated Situations
(1) Cancelling zero factors before evaluating for the
0
0
0
0
type limits When facing a
type limit, one must try to get rid of the zero factor(s) before evaluating the limit. In
0
0
is always a wrong answer. See Examples 1-3.
x
sin x
= 1 = lim
in 00 type limits involving trigono(2) Use of the basic limit lim
x→0 sin x
x→0 x
metric functions. Apply trigonometric identities and substitution to convert the current
any case,
situation into one that can apply the basic limit. See Examples 4-6.
(3) Use of Squeeze Law See Examples 7-8.
x−1
.
x→1 x2 − 1
Solution: When x → 1, both x − 1 and x2 − 1 go to 0, and so this is a
Example 1 Find lim
x2
0
0
type limit. Factor
− 1 = (x − 1)(x + 1). Then cancel the zero factors before evaluating the limit:
x−1
x−1
1
1
= lim
= lim
= .
2
x→1 x − 1
x→1 (x − 1)(x + 1)
x→1 x + 1
2
lim
√
x2 + 9 − 3
.
x→0
x2
√
Solution: When x → 0, both x2 + 9 − 3 and x2 go to 0, and so this is a
Example 2 Find lim
In order to cancel the zero factors, we need to ”free”
A2
B2
0
0
type limit.
x2
+ 9 out of the radical. Apply
√
with A = x2 + 9 and B = 3 to get
the algebraic identity (A + B)(A − B) =
−
√
√
p
( x2 + 9 − 3)( x2 + 9 + 3
x2 + 9 − 9
x2
2
√
x +9−3=
=√
=√
.
x2 + 9 + 3
x2 + 9 − 3
x2 + 9 − 3
Therefore,
√
lim
x→0
1
x2 + 9 − 3
x2
1
√
= .
=
lim
= lim √
2
2
2
2
x→0 x ( x + 9 + 3)
x→0
6
x
x +9+3
1
2
Example 3 Find lim
− 2
.
x→1 x − 1
x −1
Solution: When x → 1, both x − 1 and x2 − 1 go to 0, and so this is a ∞ − ∞ type limit.
One can use algebra to convert it into a
follows.
0
0
type limit, by combining the two fractions, as
1
2
(x + 1) − 2
x−1
−
=
= 2
.
x − 1 x2 − 1
x2 − 1
x −1
1
This becomes the problem of Example 1.
1 − cos x
.
x→0
x2
Solution: Using the algebraic identity (a − b)(a + b) = a2 − b2 , we have (1 − cos x)(1 +
Example 4 Find lim
cos x) = 1 − cos2 x; and applying a trigonometric identity sin2 x + cos2 x = 1, we have
(1 − cos x)(1 + cos x) = 1 − cos2 x = sin2 x. Thus
lim
x→0
1 − cos x
x2
=
=
=
=
=
Example 5 Find lim
x→0
1 − cos x 1 + cos x
(1 − cos x)(1 + cos x)
1
= lim
·
x2
1 + cos x x→0
x2
1 + cos x
sin2 x
1
lim
apply Limit Law (4)
x→0 x2 1 + cos x
sin x
sin x
1
lim
lim
lim
apply the Basic Limit
x→0 x x→0 x x→0 1 + cos x
1
1·1·
note that cos 0 = 1
1 + cos 0
1
.
2
lim
x→0
x
.
tan(3x)
Solution: Use substitution u = 3x. Then x = u3 , and as x → 0, we also have u → 0. Note
that tan u =
sin u
cos u .
Thus apply the Basic Limit and cos 0 = 1 to get
x
u 1
u cos u
cos u u
1
1
= lim
= lim
== lim
= ·1= .
x→0 tan(3x)
u→0 3 tan u
u→0 3 sin u
u→0 3 sin u
3
3
lim
Example 6 Find limπ (x −
x→ 2
Solution: Rewrite tan x =
π
) tan(x).
2
sin x
cos x .
Then cos π2 = 0, and so in fraction form, this limit is a
0
0 -type.
To apply the Basic Limit, we use substitution u = x − π2 . As as x → π2 , we have u → 0.
With this substitution, we have
tan(x) = tan x −
π π
+
2
2
= tan u +
π
2
=
sin(u + π2 )
.
cos(u + π2 )
Apply trigonometric identities
π
sin u +
2
π
= cos u, and cos u +
2
2
= − sin u
to get the answer:
limπ (x −
x→ 2
sin(u + π2 )
apply the trig. identities above
u→0 cos(u + π )
2
cos u
u
rewrite the expression to get
= lim u
u→0 − sin u
sin u
u
= lim − cos u
apply the Basic Limit and cos 0 = 1
u→0
sin u
= (−1) · 1 = −1.
π
) tan(x) =
2
lim u
The Squeeze Theorem states that if
f (x) ≤ g(x) ≤ h(x)
hold for all x in some interval (c, d), except possibly at the point a ∈ (c, d), and if for some
finite number L.
lim f (x) = lim h(x) = L,
x→a
x→a
then
lim g(x) = L.
x→a
Example 7 Find lim x2 sin
x→0
1
.
x
Solution: To apply Squeeze Theorem, here g(x) = x2 sin
1
. Note that −1 ≤ sin(u(x)) ≤
x
1 for any function u(x) (here u(x) = x1 ). From
1
x
−1 ≤ sin
≤ 1,
we can get, by multiplying a non negative quantity x2 everywhere in the inequalities above,
−x2 ≤ x2 sin
1
x
≤ x2 .
Hence we can choose f (x) = −x2 and h(x) = x2 . Then for the interval (−1, 1) excluding
x = 0, we have f (x) ≤ g(x) ≤ h(x). Moreover, for L = 0,
lim f (x) = lim −x2 = 0 and lim h(x) = lim x2 = 0.
x→a
x→a
x→a
It follows by the Squeeze Theorem that
lim x2 sin
1
x
x→0
3
= 0.
x→a
Example 8 Find lim
√
x→0+
x cos2
1
.
x
Solution: To apply Squeeze Theorem, here g(x) =
1]
x ).
cos(u(x)) ≤ 1 for any function u(x) (here u(x) =
0 ≤ cos2
1
x
√
x cos
2
1
.
x
Note that −1 ≤
From
≤ 1,
we can get, by multiplying a non negative quantity
0≤
√
x cos2
1
x
√
√
x everywhere in the inequalities above,
√
Hence we can choose f (x) = 0 and h(x) =
≤
x.
x. Then for the interval (0, 1), we have
f (x) ≤ g(x) ≤ h(x). Moreover, for L = 0,
lim f (x) = lim 0 = 0 and lim h(x) = lim
x→a
x→a
x→a
√
x→a
x = 0.
It follows by the Squeeze Theorem that
lim
√
x→0
x cos
2
1
x
= 0.
Example 9 Suppose that F (x) is bounded: that is, there exist constants m and n such
that m ≤ F (x) ≤ n for all x. Use the Squeeze Theorem to prove that lim x2 F (x) = 0.
x→0
2
Solution: To apply Squeeze Theorem, here g(x) = x F (x). From
m ≤ F (x) ≤ n,
we can get, by multiplying a non negative quantity x2 everywhere in the inequalities above,
−mx2 ≤ x2 F (x) ≤ nx2 .
Hence we can choose f (x) = −mx2 and h(x) = nx2 . Then for all x, we have f (x) ≤ g(x) ≤
h(x). Moreover, for L = 0,
lim f (x) = lim −mx2 = 0 and lim h(x) = lim nx2 = 0.
x→a
x→a
x→a
It follows by the Squeeze Theorem that
lim x2 F (x) = 0.
x→0
4
x→a