e) $lim_{x
ightarrow0}\frac{4 - sqrt{x^{2}+16}}{x^{2}}$

Question

Accepted Answer

# Explanation:
## Step1: Rationalize the numerator
Multiply the fraction by $\frac{4 + \sqrt{x^{2}+16}}{4 + \sqrt{x^{2}+16}}$.
$$
\begin{align*}
&\lim_{x
ightarrow0}\frac{(4 - \sqrt{x^{2}+16})(4+\sqrt{x^{2}+16})}{x^{2}(4 + \sqrt{x^{2}+16})}\
=&\lim_{x
ightarrow0}\frac{16-(x^{2}+16)}{x^{2}(4 + \sqrt{x^{2}+16})}\
=&\lim_{x
ightarrow0}\frac{16 - x^{2}-16}{x^{2}(4 + \sqrt{x^{2}+16})}\
=&\lim_{x
ightarrow0}\frac{-x^{2}}{x^{2}(4 + \sqrt{x^{2}+16})}
\end{align*}
$$
## Step2: Simplify the fraction
Cancel out the $x^{2}$ terms.
$$
\begin{align*}
&\lim_{x
ightarrow0}\frac{-x^{2}}{x^{2}(4 + \sqrt{x^{2}+16})}\
=&\lim_{x
ightarrow0}\frac{- 1}{4+\sqrt{x^{2}+16}}
\end{align*}
$$
## Step3: Evaluate the limit
Substitute $x = 0$ into the simplified function.
$$
\begin{align*}
&\frac{-1}{4+\sqrt{0^{2}+16}}\
=&\frac{-1}{4 + 4}\
=&-\frac{1}{8}
\end{align*}
$$
# Answer:
$-\frac{1}{8}$

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QUESTION IMAGE

Question

Explanation:

Step1: Rationalize the numerator

Step2: Simplify the fraction

Step3: Evaluate the limit

Answer: