# Calculate Limits of Trigonometric Functions

Several examples related to the limits of trigonometric functions with detailed solutions and exercises with answers are presented.

## Examples and Solutions

### Example 1

Find the limit

Solution to Example 1:
Let us multiply the numerator and denominator by and write

The numerator becomes is equal to , hence

The limit can be written

We have used the theorem: .



### Example 2

Find the limit $$\lim_{x \to 0} \dfrac{\sin 4 x}{4 x}$$
Solution to Example 2:
Let $$t = 4 x$$. When $$x$$ approaches 0, $$t$$ approaches 0, so that
$$\lim_{x \to 0} \dfrac {\sin 4 x}{4 x} = \lim_{t \to 0} \dfrac {\sin t}{t}$$
We now use the theorem: $$\lim_{t \to 0} \dfrac {\sin t}{t} = 1$$ to find the limit
$$\lim_{x \to 0} \dfrac {\sin 4 x}{4 x} = \lim_{t \to 0} \dfrac {\sin t}{t} = 1$$

### Example 3

Find the limit $$\lim_{t \to 0} \dfrac{\sin 6 x}{5 x}$$
Solution to Example 3:
Let $$t = 6 x$$ or $$x = t / 6$$. When $$x$$ approaches 0, $$t$$ approaches 0, so that
$$\lim_{t \to 0} \dfrac{\sin 6 x}{5 x} = \lim_{t \to 0} \dfrac{\sin t}{5 t/6}$$
$$= \lim_{t \to 0} (6 / 5) \dfrac {\sin t}{t}$$
$$= (6 / 5) \lim_{t \to 0} \dfrac {\sin t}{t}$$
$$= (6 / 5) \cdot 1 = 6 / 5$$

### Example 4

Find the limit $$\lim_{x \to -3} \dfrac{\sin (x + 3)}{x^2 +7x + 12}$$
Solution to Example 4:
If we apply the theorem of the limit of the quotient of two functions, we will get the indeterminate form $$\dfrac{0}{0}$$. We need to find another way. For $$x = -3$$, the denominator is equal to zero and therefore may be factorized, hence
$$\lim_{x \to -3} \dfrac{\sin (x + 3)}{x^2 +7x + 12}$$
$$= \lim_{x \to -3} \dfrac{\sin (x + 3)}{(x + 3)(x + 4)}$$
Let $$t = x + 3$$ or $$x = t - 3$$. As $$x$$ approaches $$-3$$, $$t$$ approaches 0.
$$\lim_{x \to -3} \dfrac{\sin (x + 3)}{x^2+7x + 12}$$
$$= \lim_{t \to 0} \dfrac {\sin t} { t (t + 1) }$$
We now apply the theorem of the limit of the product of two functions.
$$= \lim_{t \to 0} \dfrac{\sin t}{t} \cdot \lim_{t \to 0} \dfrac{1}{t+1}$$
$$= 1 \cdot 1 = 1$$

### Example 5

Find the limit $$\lim_{x \to 0} \dfrac{\sin|x|}{x}$$
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 $$x \lt 0$$, $$| x | = -x$$
$$\lim_{x \to 0^-} \dfrac{\sin|x|}{x}$$
$$= \lim_{x \to 0^-} \dfrac{\sin (- x)}{x}$$
$$= - \lim_{x \to 0^-} \dfrac{\sin x}{x}$$
$$= -1$$
For $$x \gt 0$$, $$| x | = x$$
$$\lim_{x \to 0^+} \dfrac{\sin | x |}{x}$$
$$\lim_{x \to 0^+} \dfrac{\sin x }{x}$$
$$= 1$$
The limits from the left and from the right have different values, therefore the above limit does not exist.
$$\lim_{x \to 0} \dfrac{\sin | x |}{x}$$ does not exist

### Example 6

Find the limit $$\lim_{x \to 0} \dfrac{x}{\tan x}$$
Solution to Example 6:
We first use the trigonometric identity $$\tan x = \dfrac{\sin x}{\cos x}$$
$$\lim_{x \to 0} \dfrac{x}{\tan x}$$
$$= \lim_{x \to 0} \dfrac{x}{\dfrac{\sin x}{\cos x}}$$
$$= \lim_{x \to 0} \dfrac{ x \cos x}{\sin x}$$
$$= \lim_{x \to 0} \dfrac{\cos x}{\sin x / x}$$
We now use the theorem of the limit of the quotient.
$$= \dfrac{\lim_{x \to 0} \cos x}{\lim_{x \to 0} (\sin x / x)} = 1 / 1 = 1$$

### Example 7

Find the limit $$\lim_{x \to 0} x \csc x$$
Solution to Example 7:
We first use the trigonometric identity $$\csc x = 1 / \sin x$$
$$\lim_{x \to 0} x \csc x$$
$$= \lim_{x \to 0} x / \sin x$$
$$= \lim_{x \to 0} \dfrac{1}{\sin x / x}$$
The limit of the quotient is used.
$$= 1 / 1 = 1$$

## Exercises:

Find the limits
1. $$\lim_{x \to 0} \dfrac{\sin 3 x }{\sin 8 x}$$
2. $$\lim_{x \to 0} \dfrac{\tan 3x}{x}$$
3. $$\lim_{x \to 0} \sqrt x \, \csc ( 4 \sqrt x )$$
4. $$\lim_{x \to 0} \dfrac{\sin^3 3x}{x \, \sin(x^2)}$$

Find the limits
1. 3/8
2. 3
3. 1/4
4. 27