# Calculate limits of trigonometric functions

Calculate Limits of Trigonometric
Functions
Several examples with detailed solutions
presented.
Example 1: Find the limit limx→0 (1 – cos
x) / x
Solution to Example 1:
Let us multiply the numerator and
denominator of (1 – cos x) / x by (1 + cos
x) and write
limx→0 (1 – cos x) / x
= limx→0 [ (1 – cos x) / x ] *[ (1 + cos x) / (1
+ cos x) ]
The numerator becomes 1 – cos 2 x = sin
2 x, hence
limx→0 (1 – cos x) / x
= limx→0 [ (sin 2 x) / x ] *[ 1/ (1 + cos x) ]
The limit can be written
limx→0 (1 – cos x) / x
= limx→0 [ (sin x) / x ] * limx→0 [ sin x / (1 +
cos x) ] = (1)(0/2) = 0
We have used the theorem: limx→0 [ (sin
x) / x ] = 1
Example 2: Find the limit limx→0 sin 4 x /
4x
Solution to Example 2:
Let t = 4x. When x approaches 0, t = 4x
approaches 0, so that
limx→0 sin 4 x / 4 x = limt→0 sin t / t
We now use the theorem: limt→0 sin t / t =
1 to find the limit
Find the limit limx→0 sin 4 x / 4 x = limt→0
sin t / t = 1
Example 3: Find the limit limx→0 sin 6 x /
5x
Solution to Example 3:
Let t = 6 x or x = t / 6. When x
approaches 0, t = 6 x approaches 0, so
that
limx→0 sin 6 x / 5 x = limt→0 sin t / (5 t / 6)
= limt→0 (6 / 5) sin t / t
= (6 / 5) limt→0 sin t / t
= (6 / 5) * 1 = 6 / 5
Example 4: Find the limit limx→-3 sin (x +
3) / (x 2 +7x + 12)
Solution to Example 4:
If we apply the theorem of the limit of the
quotient of two functions, we will get the
indeterminate form 0 / 0. We need to find
another way. For x = -3, the denominator
is equal to zero and therefore may be
factorized, hence
limx→-3 sin (x + 3) / (x 2 +7x + 12)
= limx→-3 sin (x + 3) / (x + 3)(x + 4)
Let t = x + 3 or x = t – 3. As x approaches
-3, t approaches 0.
limx→-3 sin (x + 3) / (x 2 +7x + 12)
= limx→-3 sin (x + 3) / (x + 3)(x + 4)
= limt→0 sin t / [ t (t + 1) ]
We now apply the theorem of the limit of
the product of two functions.
= limt→0 sin t / t * limt→0 1 / (t + 1)
= (1)*(1) = 1
Example 5: Find the limit limx→0 sin | x | /
x
Solution to Example 5:
We shall find the limit as x approaches 0
from the left and as x approaches 0 from
the right. For x 0, | x | = x
limx→0 + sin | x | / x
= limx→0 + sin x / x
=1
The two limits from the left and from the
right are different, therefore the above
limit does not exist.
limx→0 sin | x | / x does not exist
Example 6: Find the limit limx→0 x / tan x
Solution to Example 6:
We first use the trigonometric identity tan
x = sin x / cos x
= -1
limx→0 x / tan x
= limx→0 x / (sin x / cos x)
= limx→0 x cos x / sin x
= limx→0 cos x / (sin x / x)
We now use the theorem of the limit of
the quotient.
= [ limx→0 cos x ] / [ limx→0 sin x / x ] = 1 /
1=1
Example 7: Find the limit limx→0 x csc x
Solution to Example 7:
We first use the trigonometric identity csc
x = 1 / sin x
limx→0 x csc x
= limx→0 x / sin x
= limx→0 1 / (sin x / x)
The limit of the quotient is used.
=1/1=1
Exercises: Find the limits
1. limx→0 (sin 3x / sin 8x)
2. limx→0 tan 3x / x
3. limx→0 sqrt(x) csc [ 4sqrt(x) ]
4. limx→0 sin 3 3x / x sin(x 2)
Solutions to Above Exercises: Find the
limits
1. 3 / 8
2. 3
3. 1 / 4
4. 27
Calculate Limits of Trigonometric
Functions
Several examples with detailed solutions
presented.
Example 1: Find the limit limx→0 (1 – cos
x) / x
Solution to Example 1:
Let us multiply the numerator and
denominator of (1 – cos x) / x by (1 + cos
x) and write
limx→0 (1 – cos x) / x
= limx→0 [ (1 – cos x) / x ] *[ (1 + cos x) / (1
+ cos x) ]
The numerator becomes 1 – cos 2 x = sin
2 x, hence
limx→0 (1 – cos x) / x
= limx→0 [ (sin 2 x) / x ] *[ 1/ (1 + cos x) ]
The limit can be written
limx→0 (1 – cos x) / x
= limx→0 [ (sin x) / x ] * limx→0 [ sin x / (1 +
cos x) ] = (1)(0/2) = 0
We have used the theorem: limx→0 [ (sin
x) / x ] = 1
Example 2: Find the limit limx→0 sin 4 x /
4x
Solution to Example 2:
Let t = 4x. When x approaches 0, t = 4x
approaches 0, so that
limx→0 sin 4 x / 4 x = limt→0 sin t / t
We now use the theorem: limt→0 sin t / t =
1 to find the limit
Find the limit limx→0 sin 4 x / 4 x = limt→0
sin t / t = 1
Example 3: Find the limit limx→0 sin 6 x /
5x
Solution to Example 3:
Let t = 6 x or x = t / 6. When x
approaches 0, t = 6 x approaches 0, so
that
limx→0 sin 6 x / 5 x = limt→0 sin t / (5 t / 6)
= limt→0 (6 / 5) sin t / t
= (6 / 5) limt→0 sin t / t
= (6 / 5) * 1 = 6 / 5
Example 4: Find the limit limx→-3 sin (x +
3) / (x 2 +7x + 12)
Solution to Example 4:
If we apply the theorem of the limit of the
quotient of two functions, we will get the
indeterminate form 0 / 0. We need to find
another way. For x = -3, the denominator
is equal to zero and therefore may be
factorized, hence
limx→-3 sin (x + 3) / (x 2 +7x + 12)
= limx→-3 sin (x + 3) / (x + 3)(x + 4)
Let t = x + 3 or x = t – 3. As x approaches
-3, t approaches 0.
limx→-3 sin (x + 3) / (x 2 +7x + 12)
= limx→-3 sin (x + 3) / (x + 3)(x + 4)
= limt→0 sin t / [ t (t + 1) ]
We now apply the theorem of the limit of
the product of two functions.
= limt→0 sin t / t * limt→0 1 / (t + 1)
= (1)*(1) = 1
Example 5: Find the limit limx→0 sin | x | /
x
Solution to Example 5:
We shall find the limit as x approaches 0
from the left and as x approaches 0 from
the right. For x 0, | x | = x
limx→0 + sin | x | / x
= limx→0 + sin x / x
=1
The two limits from the left and from the
right are different, therefore the above
limit does not exist.
limx→0 sin | x | / x does not exist
Example 6: Find the limit limx→0 x / tan x
Solution to Example 6:
We first use the trigonometric identity tan
x = sin x / cos x
= -1
limx→0 x / tan x
= limx→0 x / (sin x / cos x)
= limx→0 x cos x / sin x
= limx→0 cos x / (sin x / x)
We now use the theorem of the limit of
the quotient.
= [ limx→0 cos x ] / [ limx→0 sin x / x ] = 1 /
1=1
Example 7: Find the limit limx→0 x csc x
Solution to Example 7:
We first use the trigonometric identity csc
x = 1 / sin x
limx→0 x csc x
= limx→0 x / sin x
= limx→0 1 / (sin x / x)
The limit of the quotient is used.
=1/1=1
Exercises: Find the limits
1. limx→0 (sin 3x / sin 8x)
2. limx→0 tan 3x / x
3. limx→0 sqrt(x) csc [ 4sqrt(x) ]
4. limx→0 sin 3 3x / x sin(x 2)
Solutions to Above Exercises: Find the
limits
1. 3 / 8
2. 3
3. 1 / 4
4. 27
v

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