Find the limit, if it exists.
-$\lim _ { - \theta } \frac { \sqrt { 1 + x } - 1 } { x }$
A) $1 / 2$
B) 0
C) Does not exist
D) $1 / 4$

Question

Quizplus · Accepted Answer

We can start by simplifying the expression in the limit. Using the conjugate, we have:

\begin{align*}
\lim_{- 	heta} \frac{\sqrt{1+x} - 1}{x} &= \lim_{- 	heta} \frac{\sqrt{1+x} - 1}{x} \cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} \
&= \lim_{- 	heta} \frac{1+x-1}{x(\sqrt{1+x}+1)} \
&= \lim_{- 	heta} \frac{x}{x(\sqrt{1+x}+1)} \
&= \lim_{- 	heta} \frac{1}{\sqrt{1+x}+1} \
&= \frac{1}{2}.
\end{align*}

Therefore, the limit is $\boxed{	ext{(A) }1/2}$.

Find the Limit, If It Exists lim⁡−θ1+x−1x\lim _ { - \theta } \frac { \sqrt { 1 + x } - 1 } { x }lim−θ​x1+x​−1​

Find the Limit, If It Exists $\lim _ { - \theta } \frac { \sqrt { 1 + x } - 1 } { x }$