## Examples with Solutions

### Example 1

Find the limit
lim_{x→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

lim_{x→0} (1 - cos x) / x

= lim_{x→0} [ (1 - cos x) / x ] *[ (1 + cos x) / (1 + cos x) ]

The numerator becomes 1 - cos ^{ 2} x = sin ^{ 2} x, hence

lim_{x→0} (1 - cos x) / x

= lim_{x→0} [ (sin ^{ 2} x) / x ] *[ 1/ (1 + cos x) ]

The limit can be written

lim_{x→0} (1 - cos x) / x

= lim_{x→0} [ (sin x) / x ] * lim_{x→0} [ sin x / (1 + cos x) ] = (1)(0/2) = 0

We have used the theorem: lim_{x→0} [ (sin x) / x ] = 1

### Example 2

Find the limit lim_{x→0} sin 4 x / 4 x
__Solution to Example 2:__

Let t = 4x. When x approaches 0, t = 4x approaches 0, so that

lim_{x→0} sin 4 x / 4 x = lim_{t→0} sin t / t

We now use the theorem: lim_{t→0} sin t / t = 1 to find the limit

Find the limit lim_{x→0} sin 4 x / 4 x = lim_{t→0} sin t / t = 1

### Example 3

Find the limit lim_{x→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

lim_{x→0} sin 6 x / 5 x = lim_{t→0} sin t / (5 t / 6)

= lim_{t→0} (6 / 5) sin t / t

= (6 / 5) lim_{t→0} sin t / t

= (6 / 5) * 1 = 6 / 5

### Example 4

Find the limit lim_{x→-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

lim_{x→-3} sin (x + 3) / (x^{ 2} +7x + 12)

= lim_{x→-3} sin (x + 3) / (x + 3)(x + 4)

Let t = x + 3 or x = t - 3. As x approaches -3, t approaches 0.

lim_{x→-3} sin (x + 3) / (x^{ 2} +7x + 12)

= lim_{x→-3} sin (x + 3) / (x + 3)(x + 4)

= lim_{t→0} sin t / [ t (t + 1) ]

We now apply the theorem of the limit of the product of two functions.

= lim_{t→0} sin t / t * lim_{t→0} 1 / (t + 1)

= (1)*(1) = 1

### Example 5

Find the limit lim_{x→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

lim_{x→0 -} sin | x | / x

= lim_{x→0 -} sin (- x ) / x

= - lim_{x→0 -} sin ( x ) / x

= -1

For x > 0, | x | = x

lim_{x→0 +} sin | x | / x

= lim_{x→0 +} sin x / x

= 1

The two limits from the left and from the right are different, therefore the above limit does not exist.

lim_{x→0} sin | x | / x does not exist

### Example 6

Find the limit lim_{x→0} x / tan x
__Solution to Example 6:__

We first use the trigonometric identity tan x = sin x / cos x

= -1

lim_{x→0} x / tan x

= lim_{x→0} x / (sin x / cos x)

= lim_{x→0} x cos x / sin x

= lim_{x→0} cos x / (sin x / x)

We now use the theorem of the limit of the quotient.

= [ lim_{x→0} cos x ] / [ lim_{x→0} sin x / x ] = 1 / 1 = 1

### Example 7

Find the limit lim_{x→0} x csc x
__Solution to Example 7:__

We first use the trigonometric identity csc x = 1 / sin x

lim_{x→0} x csc x

= lim_{x→0} x / sin x

= lim_{x→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})

### ExercisesSolutions to Above Exercises

1. 3 / 8

2. 3

3. 1 / 4

4. 27
## More Links on limits

Calculus Tutorials and Problems

Limits of Absolute Value Functions Questions