what is $lim_{x
ightarrow2}\frac{sqrt{x^{2}+5}-3}{x^{2}-4}$ equal to?

Question

Accepted Answer

# Explanation:
## Step1: Rationalize the numerator
Multiply the fraction by $\frac{\sqrt{x^{2}+5}+3}{\sqrt{x^{2}+5}+3}$.
$$
\begin{align*}
&\lim_{x
ightarrow2}\frac{\sqrt{x^{2}+5}-3}{x^{2}-4}\times\frac{\sqrt{x^{2}+5}+3}{\sqrt{x^{2}+5}+3}\
=&\lim_{x
ightarrow2}\frac{(x^{2}+5 - 9)}{(x^{2}-4)(\sqrt{x^{2}+5}+3)}\
=&\lim_{x
ightarrow2}\frac{x^{2}-4}{(x^{2}-4)(\sqrt{x^{2}+5}+3)}
\end{align*}
$$
## Step2: Simplify the fraction
Cancel out the common factor $(x^{2}-4)$ (since $x
eq2$ when taking the limit).
$$
\begin{align*}
&\lim_{x
ightarrow2}\frac{x^{2}-4}{(x^{2}-4)(\sqrt{x^{2}+5}+3)}\
=&\lim_{x
ightarrow2}\frac{1}{\sqrt{x^{2}+5}+3}
\end{align*}
$$
## Step3: Substitute $x = 2$
$$
\begin{align*}
&\frac{1}{\sqrt{2^{2}+5}+3}\
=&\frac{1}{\sqrt{9}+3}\
=&\frac{1}{3 + 3}\
=&\frac{1}{6}
\end{align*}
$$
# Answer:
$\frac{1}{6}$

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QUESTION IMAGE

Question

Explanation:

Step1: Rationalize the numerator

Step2: Simplify the fraction

Step3: Substitute $x = 2$

Answer: