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…

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

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

Answer

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

Step3: Use the limit - property

Let (t=x + 8), when (x\rightarrow - 8), (t\rightarrow0). Then (\lim_{x\rightarrow - 8}\frac{\sin(x + 8)}{(x + 8)(x + 3)}=\lim_{t\rightarrow0}\frac{\sin t}{t(t - 5)})

We know that (\lim_{t\rightarrow0}\frac{\sin t}{t}=1)

(\lim_{t\rightarrow0}\frac{\sin t}{t(t - 5)}=\lim_{t\rightarrow0}\frac{\sin t}{t}\cdot\lim_{t\rightarrow0}\frac{1}{t - 5})

Step4: Evaluate the limits

(\lim_{t\rightarrow0}\frac{\sin t}{t}=1) and (\lim_{t\rightarrow0}\frac{1}{t - 5}=\frac{1}{0 - 5}=-\frac{1}{5})

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

Answer:

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