Evaluating Trigonometric Limits Analytically

The limit we just proved above can help us evaluate many related trigonometric limits that initially have the indeterminate form $\dfrac{0}{0}.$

In Example $\PageIndex{11}$, we use this limit to establish $\dlim_{θ→0}\dfrac{1−\cos θ}{θ}=0$. This limit also proves useful in later chapters.

Example $\PageIndex{11}$: Evaluating an Important Trigonometric Limit

Evaluate $\dlim_{θ→0}\dfrac{1−\cos θ}{θ}$.

Solution

In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine:

$$\begin{align*} \lim_{θ→0}\dfrac{1−\cos θ}{θ} &=\dlim_{θ→0}\dfrac{1−\cos θ}{θ}⋅\dfrac{1+\cos θ}{1+\cos θ}\\[4pt] &= \lim_{θ→0}\dfrac{1−\cos^2θ}{θ(1+\cos θ)}\\[4pt] &= \lim_{θ→0}\dfrac{\sin^2θ}{θ(1+\cos θ)}\\[4pt] &= \lim_{θ→0}\dfrac{\sin θ}{θ}⋅\dfrac{\sin θ}{1+\cos θ}\\[4pt] &= \left(\lim_{θ→0}\dfrac{\sin θ}{θ} \right)\cdot\left( \lim_{θ→0} \dfrac{\sin θ}{1+\cos θ}\right) \\[4pt] &= 1⋅\dfrac{0}{2}=0. \end{align*}$$

Therefore,

$$\lim_{θ→0}\dfrac{1−\cos θ}{θ}=0.$$

Exercise $\PageIndex{11}$

Evaluate $\dlim_{θ→0}\dfrac{1−\cos θ}{\sin θ}$.

Hint

Multiply numerator and denominator by $1+\cos θ$.

Answer

$\displaystyle \begin{align*} \lim_{θ→0}\dfrac{1−\cos θ}{\sin θ} &= \lim_{θ→0}\dfrac{1−\cos θ}{\sin θ}\cdot \dfrac{1+\cos θ}{1+\cos θ}\\[5pt] &= \lim_{θ→0}\dfrac{1−\cos^2 θ}{\big(\sin θ\big)\big( 1+\cos θ \big)} \\[5pt] &= \lim_{θ→0}\dfrac{\sin^2 θ}{\big(\sin θ\big)\big( 1+\cos θ \big)} \\[5pt] &= \lim_{θ→0}\dfrac{\sin θ}{1+\cos θ} \\[5pt] &= \dfrac{\sin 0}{1 + \cos 0} = \dfrac{0}{1+1} =\dfrac{0}{2} \\[5pt] &= 0 \end{align*}$

Example $\PageIndex{12}$

a. Evaluate $\dlim_{θ→0}\dfrac{5θ^3+\sin^5 θ}{θ^3}$.

b. Evaluate $\dlim_{x→0}\dfrac{\tan^2 (7x)}{5x}$.

c. Evaluate $\dlim_{x→5}\dfrac{\sin (x-5)}{x^2-25}$.

Solution

In all three of these examples, we notice that plugging in the target value gives us the indeterminate form $\dfrac{0}{0}.$  That means there is likely a way to evaluate the limit, and since these limits all can be written in terms of a sine function, we attempt to rewrite the limit expression in such a way that we can use the limit we proved above: $$\lim_{θ→0}\dfrac{\sin θ}{θ}=1.\nonumber$$

a.  Evaluate $\dlim_{θ→0}\dfrac{5θ^3+\sin^5 θ}{θ^3}$.
$\displaystyle \begin{align*} \lim_{θ→0}\dfrac{5θ^3+\sin^5 θ}{θ^3} &= \lim_{θ→0}\left[\dfrac{5θ^3}{θ^3}+\dfrac{\sin^3 θ}{θ^3}\cdot \sin^2 θ \right] \\[5pt] &= \lim_{θ→0}\left[5+\left(\dfrac{\sin θ}{θ}\right)^3\cdot \sin^2 θ \right] \\[5pt] &= \lim_{θ→0} 5 + \lim_{θ→0}\left(\dfrac{\sin θ}{θ}\right)^3\cdot \lim_{θ→0} \sin^2 θ & & \text{Using the Sum and Product Rules.}\\[5pt] &= 5 + \left(\lim_{θ→0}\dfrac{\sin θ}{θ}\right)^3\cdot \left(\lim_{θ→0}\sin θ\right)^2 & & \text{Using the Constant Rule and Power Rule.}\\[5pt] &= 5 + 1^3 \cdot 0^2 & & \text{Using limits proven above.} \\[5pt] &=5 \end{align*}$

Note how we needed to use the limit rules to rewrite this limit so that the limit, $\dlim_{θ→0}\dfrac{\sin θ}{θ}$ was set off by itself.  Once we have that limit set off, we know its value from what we proved earlier and we were able to evaluate the limit clearly and without issues.

 

b. Evaluate $\dlim_{x→0}\dfrac{\tan^2 (7x)}{5x}$.
$\displaystyle \begin{align*} \lim_{x→0}\dfrac{\tan^2 (7x)}{5x} &= \lim_{x→0}\left[\dfrac{\sin(7x)}{\cos(7x)}\cdot\dfrac{\tan(7x)}{5x} \right] \\[5pt] &= \lim_{x→0}\left[\dfrac{\sin(7x)}{7x}\cdot\dfrac{7\tan(7x)}{5\cos(7x)} \right] & & \text{We rewrite and multiply through by }\frac{7}{7}\text{ to obtain form }\frac{\sin u}{u}\text{ with } u = 7x.\\[5pt] &= \left(\lim_{x→0}\dfrac{\sin (7x)}{7x}\right)\cdot \frac{7}{5}\lim_{x→0} \dfrac{\tan(7x)}{\cos(7x)} & & \text{We set off }\lim_{x→0}\dfrac{\sin (7x)}{7x}\text{ by itself.}\\[5pt] &= 1\cdot\dfrac{7}{5}\cdot\dfrac{\tan 0}{\cos 0} & & \text{Now each part can be evaluated.}\\[5pt] &= \dfrac{7}{5}\cdot\dfrac{0}{1}\\[5pt] &= \dfrac{7}{5}\cdot 0\\[5pt] &=0 \end{align*}$

 

c. Evaluate $\dlim_{x→5}\dfrac{\sin (x-5)}{x^2-25}$.

This example is the most interesting and requires a little more insight.  We observe that there is a sine function in the limit expression and we see that it does have the form $\frac{0}{0},$ so we will seek to rewrite it so that part of the limit can be expressed in terms of $\dlim_{u→0} \frac{\sin u}{u}$ for some $u$.

$\displaystyle \begin{align*} \lim_{x→5}\dfrac{\sin (x-5)}{x^2-25} &= \lim_{x→5}\dfrac{\sin (x-5)}{x-5}\cdot\dfrac{1}{x+5} & & \text{Here we factored the denominator and see that }u=x-5 \text{ is a natural choice.}\\[5pt] &= \lim_{x→5}\dfrac{\sin (x-5)}{x-5}\cdot\lim_{x→5}\dfrac{1}{x+5} \end{align*}$

Now, is we let  $u = x - 5,$ then as $x \to 5, \; u \to 0,$ so we could rewrite $\dlim_{x→5}\dfrac{\sin (x-5)}{x-5} = \lim_{u→0}\dfrac{\sin u}{u}.$

Then we have:

$\displaystyle \begin{align*} \lim_{x→5}\dfrac{\sin (x-5)}{x^2-25} &= \lim_{x→5}\dfrac{\sin (x-5)}{x-5}\cdot\lim_{x→5}\dfrac{1}{x+5}\\[5pt] &=\left(\lim_{u→0}\dfrac{\sin u}{u}\right)\cdot\left(\dfrac{1}{5+5}\right) & & \text{We can evaluate the second limit by direct substitution.} \\[5pt] &= 1 \cdot \frac{1}{10}\\[5pt] &= \frac{1}{10}\end{align*}$

Contributors:

Example 11 and the statement of Exercise 11 is from the original OpenStax Calculus text.

The rest, including the solution to Exercise 11 is by Paul Seeburger (Monroe Community College).