# Calculate Limits of Trigonometric Functions

Several examples on how to find limits of trigonometric functions, with detailed solutions, and exercises with answers are presented.

## Examples with Solutions

### 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 / 4 x

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 / 5 x

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
= - limx→0 - sin ( x ) / x
= -1
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. lim
x→0 (sin 3x / sin 8x)
2. lim
x→0 tan 3x / x
3. lim
x→0 sqrt(x) csc [ 4sqrt(x) ]
4. lim
x→0 sin 3 3x / x sin(x 2)

1. 3 / 8
2. 3
3. 1 / 4
4. 27