Question

$$lim_{x ightarrow0}\frac{sqrt{1 + x}-1}{x}$$

Explanation:

Step1: Rationalize the numerator

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

$$\begin{align*} &\lim_{x ightarrow0}\frac{\sqrt{1 + x}-1}{x}\times\frac{\sqrt{1 + x}+1}{\sqrt{1 + x}+1}\\ =&\lim_{x ightarrow0}\frac{(1 + x)-1}{x(\sqrt{1 + x}+1)}\\ =&\lim_{x ightarrow0}\frac{x}{x(\sqrt{1 + x}+1)} \end{align*}$$

\]

Step2: Simplify the fraction

Cancel out the common factor $x$ in the numerator and denominator.
\[

$$\begin{align*} &\lim_{x ightarrow0}\frac{x}{x(\sqrt{1 + x}+1)}\\ =&\lim_{x ightarrow0}\frac{1}{\sqrt{1 + x}+1} \end{align*}$$

\]

Step3: Substitute $x = 0$

\[

$$\begin{align*} &\frac{1}{\sqrt{1+0}+1}\\ =&\frac{1}{1 + 1}\\ =&\frac{1}{2} \end{align*}$$

\]

Answer:

$\frac{1}{2}$