Find $\lim _ { x \rightarrow 0 ^ { - } } \frac { x } { \sqrt { x + 3 } - \sqrt { 3 } }$
A) 6
B) $\frac { 3 } { 2 }$
C) $\frac { \sqrt { 3 } } { 2 }$
D) $2 \sqrt { 3 }$
E) limit does not exist

Question

Quizplus · Accepted Answer

Multiplying the numerator and denominator by the conjugate gives
\begin{align*}
\lim_{x	o 0^-}\frac{x(\sqrt{x+3}+\sqrt{3})}{x+3-3} &= \lim_{x	o 0^-}\frac{x(\sqrt{x+3}+\sqrt{3})}{x}\
&=\lim_{x	o 0^-} (\sqrt{x+3}+\sqrt{3})\
&=\sqrt{3}+ \lim_{x	o 0^-} (\sqrt{x+3})\
&=2\sqrt{3}.
\end{align*}

Find lim⁡x→0−xx+3−3\lim _ { x \rightarrow 0 ^ { - } } \frac { x } { \sqrt { x + 3 } - \sqrt { 3 } }limx→0−​x+3​−3​x​

Find $\lim _ { x \rightarrow 0 ^ { - } } \frac { x } { \sqrt { x + 3 } - \sqrt { 3 } }$