Question

e) $lim_{x
ightarrow - 1}\frac{sqrt{x + 5}-2}{x + 1}$

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{x + 5}+2}{\sqrt{x + 5}+2}$.
\[

$$\begin{align*} &\lim_{x ightarrow - 1}\frac{\sqrt{x + 5}-2}{x + 1}\times\frac{\sqrt{x + 5}+2}{\sqrt{x + 5}+2}\\ =&\lim_{x ightarrow - 1}\frac{(\sqrt{x + 5})^2-2^2}{(x + 1)(\sqrt{x + 5}+2)}\\ =&\lim_{x ightarrow - 1}\frac{x + 5-4}{(x + 1)(\sqrt{x + 5}+2)}\\ =&\lim_{x ightarrow - 1}\frac{x + 1}{(x + 1)(\sqrt{x + 5}+2)} \end{align*}$$

\]

Step2: Simplify the fraction

Cancel out the common factor $(x + 1)$ in the numerator and denominator.
\[

$$\begin{align*} &\lim_{x ightarrow - 1}\frac{x + 1}{(x + 1)(\sqrt{x + 5}+2)}\\ =&\lim_{x ightarrow - 1}\frac{1}{\sqrt{x + 5}+2} \end{align*}$$

\]

Step3: Substitute $x=-1$

\[

$$\begin{align*} &\frac{1}{\sqrt{-1 + 5}+2}\\ =&\frac{1}{\sqrt{4}+2}\\ =&\frac{1}{2 + 2}\\ =&\frac{1}{4} \end{align*}$$

\]

Answer:

$\frac{1}{4}$