Question

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

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:

$-\frac{4}{5}$