Question

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

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:

$-\frac{1}{8}$