Question

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

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:

$\frac{1}{6}$