Several examples related to the limits of trigonometric functions with detailed solutions and exercises with answers are presented.
In what follows, \( C \) represents the constant of integration where applicable.
Examples and Solutions
Click on each example to view its detailed step-by-step solution.
Example 1
Find the limit:
\[ \lim_{x \to 0} \dfrac{1 - \cos x}{x} \]Show Solution to Example 1
Let us multiply the numerator and denominator by \( 1 + \cos x \) and write:
\[ \lim_{x \to 0} \dfrac{1 - \cos x}{x} = \lim_{x \to 0} \dfrac{1 - \cos x}{x} \cdot \dfrac{1 + \cos x}{1 + \cos x} \]The numerator becomes equal to \( 1 - \cos^2 x = \sin^2 x \), hence:
\[ \lim_{x \to 0} \dfrac{1 - \cos x}{x} = \lim_{x \to 0} \dfrac{\sin^2 x}{x} \cdot \dfrac{1}{1 + \cos x} \]The limit can be written as:
\[ \lim_{x \to 0} \dfrac{1 - \cos x}{x} = \lim_{x \to 0} \left( \dfrac{\sin x}{x} \right) \cdot \lim_{x \to 0} \left( \dfrac{\sin x}{1 + \cos x} \right) = (1) \left(\dfrac{0}{2}\right) = 0 \]We have used the theorem: \( \lim_{x \to 0} \dfrac{\sin x}{x} = 1 \).
Example 2
Find the limit:
\[ \lim_{x \to 0} \dfrac{\sin 4x}{4x} \]Show Solution to Example 2
Let \( t = 4x \). When \( x \) approaches 0, \( t \) approaches 0, so that:
\[ \lim_{x \to 0} \dfrac{\sin 4x}{4x} = \lim_{t \to 0} \dfrac{\sin t}{t} \]We now use the theorem \( \lim_{t \to 0} \dfrac{\sin t}{t} = 1 \) to find the limit:
\[ \lim_{x \to 0} \dfrac{\sin 4x}{4x} = \lim_{t \to 0} \dfrac{\sin t}{t} = 1 \]Example 3
Find the limit:
\[ \lim_{t \to 0} \dfrac{\sin 6x}{5x} \]Show Solution to Example 3
Let \( t = 6x \), or \( x = t / 6 \). When \( x \) approaches 0, \( t \) approaches 0, so that:
\[ \lim_{t \to 0} \dfrac{\sin 6x}{5x} = \lim_{t \to 0} \dfrac{\sin t}{5t/6} \] \[ = \lim_{t \to 0} \left(\dfrac{6}{5}\right) \dfrac{\sin t}{t} \] \[ = \left(\dfrac{6}{5}\right) \lim_{t \to 0} \dfrac{\sin t}{t} \] \[ = \left(\dfrac{6}{5}\right) \cdot 1 = \dfrac{6}{5} \]Example 4
Find the limit:
\[ \lim_{x \to -3} \dfrac{\sin (x + 3)}{x^2 + 7x + 12} \]Show 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:
\[ = \lim_{t \to 0} \dfrac{\sin t}{t(t + 1)} \]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:
\[ \lim_{x \to 0} \dfrac{\sin|x|}{x} \]Show 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 \( x < 0 \), \( |x| = -x \):
\[ \lim_{x \to 0^-} \dfrac{\sin|x|}{x} = \lim_{x \to 0^-} \dfrac{\sin(-x)}{x} = -\lim_{x \to 0^-} \dfrac{\sin x}{x} = -1 \]For \( x > 0 \), \( |x| = x \):
\[ \lim_{x \to 0^+} \dfrac{\sin|x|}{x} = \lim_{x \to 0^+} \dfrac{\sin x}{x} = 1 \]The limits from the left and from the right have different values, therefore the above limit does not exist:
\[ \lim_{x \to 0} \dfrac{\sin|x|}{x} \quad \text{DOES NOT EXIST} \]Example 6
Find the limit:
\[ \lim_{x \to 0} \dfrac{x}{\tan x} \]Show Solution to Example 6
We first use the trigonometric identity \( \tan x = \dfrac{\sin x}{\cos x} \):
\[ \lim_{x \to 0} \dfrac{x}{\tan x} = \lim_{x \to 0} \dfrac{x}{\dfrac{\sin x}{\cos x}} = \lim_{x \to 0} \dfrac{x \cos x}{\sin x} = \lim_{x \to 0} \dfrac{\cos x}{\dfrac{\sin x}{x}} \]We now use the theorem of the limit of the quotient:
\[ = \dfrac{\lim_{x \to 0} \cos x}{\lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)} = \dfrac{1}{1} = 1 \]Example 7
Find the limit:
\[ \lim_{x \to 0} x \csc x \]Show Solution to Example 7
We first use the trigonometric identity \( \csc x = \dfrac{1}{\sin x} \):
\[ \lim_{x \to 0} x \csc x = \lim_{x \to 0} \dfrac{x}{\sin x} = \lim_{x \to 0} \dfrac{1}{\dfrac{\sin x}{x}} \]The limit of the quotient is used:
\[ = \dfrac{1}{1} = 1 \]Exercises
Calculate the following limits. Click each exercise to check your answers.
Exercise 1
\( \displaystyle \lim_{x \to 0} \dfrac{\sin 3x}{\sin 8x} \)
Show Answer
Exercise 2
\( \displaystyle \lim_{x \to 0} \dfrac{\tan 3x}{x} \)
Show Answer
Exercise 3
\( \displaystyle \lim_{x \to 0} \sqrt{x} \, \csc (4 \sqrt{x}) \)
Show Answer
Exercise 4
\( \displaystyle \lim_{x \to 0} \dfrac{\sin^3 3x}{x \sin(x^2)} \)