# Calculate Limits of Trigonometric Functions

 Several examples with detailed solutions and exercises with answers are presented. Example 1: Find the limit $\lim_{x \to 0} \dfrac{1-\cos x}{x}$ Solution to Example 1: Let us multiply the numerator and denominator by $1 + \cos x$ and write $\lim_{x \to 0} \dfrac{1-\cos x}{x}$ $= \lim_{x \to 0} \dfrac{1-\cos x}{x} \dfrac{1+\cos x}{1+\cos x}$ The numerator becomes $1 - \cos^2 x = \sin^2 x$, hence $\lim_{x \to 0} \dfrac{1-\cos x}{x}$ $=\lim_{x \to 0} \dfrac{\sin^2 x}{x} \dfrac{1}{1+\cos x}$ The limit can be written $\lim_{x \to 0} \dfrac{1-\cos x}{x}$ $= \lim_{x \to 0} (\dfrac{\sin x}{x}) \cdot \lim_{x \to 0}(\dfrac{\sin x}{1+\cos x}) = (1)(0/2) = 0$ We have used the theorem: $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$ Example 2: Find the limit $\lim_{x \to 0} \dfrac{\sin 4 x}{4 x}$ 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 $x < 0$, $| x | = -x$ $\lim_{x \to 0^-} \dfrac{\sin|x|}{x}$ $= \lim_{x \to 0^-} \dfrac{\sin (- x)}{x}$ $= - \lim_{x \to 0^-} \dfrac{\sin x}{x}$ $= -1$ 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. $\lim_{x \to 0} \dfrac{\sin | x |}{x}$ 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. $\lim_{x \to 0} \dfrac{\sin^3 3x}{x \, sin(x^2)}$ Solutions to Above Exercises: Find the limits 1. $3 / 8$ 2. $3$ 3. $1 / 4$ 4. $27$ More on limits Calculus Tutorials and Problems

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Updated: 2 April 2013