Answer
$\dfrac{2}{c}$
Work Step by Step
We have to determine $L=\lim\limits_{x \to 0} \dfrac{x}{\sqrt{cx+1}-1}$.
Multiply both numerator and denominator by the conjugate of the denominator:
$L=\lim\limits_{x \to 0}\dfrac{x}{\sqrt{cx+1}-1}\cdot\dfrac{\sqrt{cx+1}+1}{\sqrt{cx+1}+1}$
$=\lim\limits_{x \to 0}\dfrac{x(\sqrt{cx+1}+1)}{(\sqrt{cx+1})^2-1^2}=\lim\limits_{x \to 0}\dfrac{x(\sqrt{cx+1}+1)}{cx+1-1}$
$=\lim\limits_{x \to 0}\dfrac{x(\sqrt{cx+1}+1)}{cx}$
Simplify:
$L=\lim\limits_{x \to 0}\dfrac{\sqrt{cx+1}+1}{c}$
Determine the limit:
$L=\dfrac{\sqrt{c(0)+1}+1}{c}=\dfrac{2}{c}$
