Several examples related to the limits of trigonometric functions with detailed solutions and exercises with answers are presented.
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: .
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
\( \lim_{x \to 0} \dfrac {\sin 4 x}{4 x} = \lim_{t \to 0} \dfrac {\sin t}{t} = 1 \)
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
\( \lim_{t \to 0} \dfrac{\sin 6 x}{5 x} = \lim_{t \to 0} \dfrac{\sin t}{5 t/6} \)
\( = \lim_{t \to 0} (6 / 5) \dfrac {\sin t}{t} \)
\( = (6 / 5) \lim_{t \to 0} \dfrac {\sin t}{t} \)
\( = (6 / 5) \cdot 1 = 6 / 5 \)
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
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.
\( = \dfrac{\lim_{x \to 0} \cos x}{\lim_{x \to 0} (\sin x / x)} = 1 / 1 = 1 \)
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.
\( = 1 / 1 = 1 \)