Question

use $lim_{x \to 0}\frac{sin x}{x}=1$ and/or $lim_{x \to 0}\frac{cos x - 1}{x}=0$ to evaluate the following limit $lim_{\theta \to 0}\frac{cos^{2}\theta - 1}{\theta}$. select the correct choice, and, if necessary, fill in the answer box to complete your choice.
a. $lim_{\theta \to 0}\frac{cos^{2}\theta - 1}{\theta}=$ (type an integer or a simplified fraction )
b. the limit is undefined.

Explanation:

Step1: Use the identity $a^2 - b^2=(a + b)(a - b)$

We know that $\cos^{2}\theta-1=(\cos\theta + 1)(\cos\theta - 1)$. So, $\lim_{\theta
ightarrow0}\frac{\cos^{2}\theta - 1}{\theta}=\lim_{\theta
ightarrow0}\frac{(\cos\theta + 1)(\cos\theta - 1)}{\theta}$.

Step2: Split the limit

By the product - rule of limits $\lim_{x
ightarrow a}(f(x)g(x))=\lim_{x
ightarrow a}f(x)\cdot\lim_{x
ightarrow a}g(x)$, we have $\lim_{\theta
ightarrow0}\frac{(\cos\theta + 1)(\cos\theta - 1)}{\theta}=\lim_{\theta
ightarrow0}(\cos\theta + 1)\cdot\lim_{\theta
ightarrow0}\frac{\cos\theta - 1}{\theta}$.

Step3: Evaluate each limit

We know that $\lim_{\theta
ightarrow0}\cos\theta=1$, so $\lim_{\theta
ightarrow0}(\cos\theta + 1)=1 + 1=2$, and $\lim_{\theta
ightarrow0}\frac{\cos\theta - 1}{\theta}=0$. Then $\lim_{\theta
ightarrow0}(\cos\theta + 1)\cdot\lim_{\theta
ightarrow0}\frac{\cos\theta - 1}{\theta}=2\times0 = 0$.

Answer:

A. $\lim_{\theta
ightarrow0}\frac{\cos^{2}\theta - 1}{\theta}=0$