Question

use $lim_{x \to 0}\frac{sin x}{x}=1$ and/or $lim_{t \to 0}\frac{cos x - 1}{x}=0$ to evaluate the following limit. $lim_{x \to - 8}\frac{sin(x + 8)}{x^{2}+11x + 24}$ select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. $lim_{x \to - 8}\frac{sin(x + 8)}{x^{2}+11x + 24}=$ (type an integer or a simplified fraction) b. the limit is undefined

Explanation:

Step1: Factor the denominator

Factor \(x^{2}+11x + 24\). We have \(x^{2}+11x + 24=(x + 8)(x+3)\)

Step2: Rewrite the limit

\(\lim_{x
ightarrow - 8}\frac{\sin(x + 8)}{x^{2}+11x + 24}=\lim_{x
ightarrow - 8}\frac{\sin(x + 8)}{(x + 8)(x + 3)}\)

Step3: Use the limit - property

Let \(t=x + 8\), when \(x
ightarrow - 8\), \(t
ightarrow0\). Then \(\lim_{x
ightarrow - 8}\frac{\sin(x + 8)}{(x + 8)(x + 3)}=\lim_{t
ightarrow0}\frac{\sin t}{t(t - 5)}\)

We know that \(\lim_{t
ightarrow0}\frac{\sin t}{t}=1\)

\(\lim_{t
ightarrow0}\frac{\sin t}{t(t - 5)}=\lim_{t
ightarrow0}\frac{\sin t}{t}\cdot\lim_{t
ightarrow0}\frac{1}{t - 5}\)

Step4: Evaluate the limits

\(\lim_{t
ightarrow0}\frac{\sin t}{t}=1\) and \(\lim_{t
ightarrow0}\frac{1}{t - 5}=\frac{1}{0 - 5}=-\frac{1}{5}\)

So \(\lim_{x
ightarrow - 8}\frac{\sin(x + 8)}{x^{2}+11x + 24}=1\times(-\frac{1}{5})=-\frac{1}{5}\)

Answer:

A. \(\lim_{x
ightarrow - 8}\frac{\sin(x + 8)}{x^{2}+11x + 24}=-\frac{1}{5}\)