Question

$lim_{x
ightarrow3}\frac{sqrt{x + 1}-2}{x - 3}$

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{x + 1}+2}{\sqrt{x + 1}+2}$.
\[

$$\begin{align*} &\lim_{x ightarrow3}\frac{\sqrt{x + 1}-2}{x - 3}\times\frac{\sqrt{x + 1}+2}{\sqrt{x + 1}+2}\\ =&\lim_{x ightarrow3}\frac{(\sqrt{x + 1})^2-2^2}{(x - 3)(\sqrt{x + 1}+2)}\\ =&\lim_{x ightarrow3}\frac{x + 1-4}{(x - 3)(\sqrt{x + 1}+2)}\\ =&\lim_{x ightarrow3}\frac{x - 3}{(x - 3)(\sqrt{x + 1}+2)} \end{align*}$$

\]

Step2: Simplify the fraction

Cancel out the common factor $(x - 3)$ in the numerator and denominator.
\[

$$\begin{align*} &\lim_{x ightarrow3}\frac{x - 3}{(x - 3)(\sqrt{x + 1}+2)}\\ =&\lim_{x ightarrow3}\frac{1}{\sqrt{x + 1}+2} \end{align*}$$

\]

Step3: Substitute $x = 3$

\[

$$\begin{align*} &\frac{1}{\sqrt{3+1}+2}\\ =&\frac{1}{\sqrt{4}+2}\\ =&\frac{1}{2 + 2}\\ =&\frac{1}{4} \end{align*}$$

\]

Answer:

$\frac{1}{4}$