Calculate Limits of Trigonometric Functions

Several examples with detailed solutions and exercises with answers are 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 / 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. 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


More on limits

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Updated: 2 April 2013

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