d. $lim_{x o - 4}\frac{sqrt{x^{2}+9}-5}{x + 4}$

Question

d. $lim_{x 	o - 4}\frac{sqrt{x^{2}+9}-5}{x + 4}$

Accepted Answer

# Explanation:
## Step1: Rationalize the numerator
Multiply the fraction by $\frac{\sqrt{x^{2}+9}+5}{\sqrt{x^{2}+9}+5}$.
$$
\begin{align*}
&\lim_{x
ightarrow - 4}\frac{\sqrt{x^{2}+9}-5}{x + 4}\times\frac{\sqrt{x^{2}+9}+5}{\sqrt{x^{2}+9}+5}\
=&\lim_{x
ightarrow - 4}\frac{(x^{2}+9)-25}{(x + 4)(\sqrt{x^{2}+9}+5)}\
=&\lim_{x
ightarrow - 4}\frac{x^{2}-16}{(x + 4)(\sqrt{x^{2}+9}+5)}
\end{align*}
$$
## Step2: Factor the numerator
Factor $x^{2}-16$ as $(x + 4)(x - 4)$.
$$
\begin{align*}
&\lim_{x
ightarrow - 4}\frac{(x + 4)(x - 4)}{(x + 4)(\sqrt{x^{2}+9}+5)}\
=&\lim_{x
ightarrow - 4}\frac{x - 4}{\sqrt{x^{2}+9}+5}
\end{align*}
$$
## Step3: Substitute $x=-4$
$$
\begin{align*}
&\frac{-4-4}{\sqrt{(-4)^{2}+9}+5}\
=&\frac{-8}{\sqrt{16 + 9}+5}\
=&\frac{-8}{\sqrt{25}+5}\
=&\frac{-8}{5 + 5}\
=&-\frac{4}{5}
\end{align*}
$$
# Answer:
$-\frac{4}{5}$

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QUESTION IMAGE

Question

Explanation:

Step1: Rationalize the numerator

Step2: Factor the numerator

Step3: Substitute $x=-4$

Answer: