Question

find the limit.
lim(x→ - 5) (4 - √(x² - 9))/(x + 5)
select the correct choice below and, if necessary, fill in the answer box to complete
a. lim(x→ - 5) (4 - √(x² - 9))/(x + 5)=5/4 (type an integer or a simplified fraction.)
b. the limit does not exist.

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{4 + \sqrt{x^{2}-9}}{4+\sqrt{x^{2}-9}}$.
\[

$$\begin{align*} &\lim_{x ightarrow - 5}\frac{4-\sqrt{x^{2}-9}}{x + 5}\times\frac{4+\sqrt{x^{2}-9}}{4+\sqrt{x^{2}-9}}\\ =&\lim_{x ightarrow - 5}\frac{16-(x^{2}-9)}{(x + 5)(4+\sqrt{x^{2}-9})}\\ =&\lim_{x ightarrow - 5}\frac{16 - x^{2}+9}{(x + 5)(4+\sqrt{x^{2}-9})}\\ =&\lim_{x ightarrow - 5}\frac{25 - x^{2}}{(x + 5)(4+\sqrt{x^{2}-9})} \end{align*}$$

\]

Step2: Factor the numerator

Factor $25 - x^{2}$ as $(5 - x)(5 + x)$ using the difference - of - squares formula $a^{2}-b^{2}=(a - b)(a + b)$ where $a = 5$ and $b=x$.
\[

$$\begin{align*} &\lim_{x ightarrow - 5}\frac{(5 - x)(5 + x)}{(x + 5)(4+\sqrt{x^{2}-9})}\\ =&\lim_{x ightarrow - 5}\frac{5 - x}{4+\sqrt{x^{2}-9}} \end{align*}$$

\]

Step3: Substitute $x=-5$

\[

$$\begin{align*} &\frac{5-(-5)}{4+\sqrt{(-5)^{2}-9}}\\ =&\frac{5 + 5}{4+\sqrt{25 - 9}}\\ =&\frac{10}{4+\sqrt{16}}\\ =&\frac{10}{4 + 4}\\ =&\frac{10}{8}\\ =&\frac{5}{4} \end{align*}$$

\]

Answer:

A. $\lim_{x
ightarrow - 5}\frac{4-\sqrt{x^{2}-9}}{x + 5}=\frac{5}{4}$