Answer
$\dfrac{3}{2}$
Work Step by Step
We have to determine $L=\lim\limits_{x \to 1} \dfrac{x-1}{\sqrt{4x+5}-3}$.
Multiply both numerator and denominator by the conjugate of the denominator:
$L=\lim\limits_{x \to 1} \dfrac{x-1}{\sqrt{4x+5}-3}\cdot\dfrac{\sqrt{4x+5}+3}{\sqrt{4x+5}+3}$
$=\lim\limits_{x \to 1} \dfrac{(x-1)(\sqrt{4x+5}+3)}{(\sqrt{4x+5})^2-3^2}$
$=\lim\limits_{x \to 1} \dfrac{(x-1)(\sqrt{4x+5}+3)}{4x+5-9}$
$=\lim\limits_{x \to 1} \dfrac{(x-1)(\sqrt{4x+5}+3)}{4x-4}$
$=\lim\limits_{x \to 1} \dfrac{(x-1)(\sqrt{4x+5}+3)}{4(x-1)}$
Simplify:
$L=\lim\limits_{x \to 1} \dfrac{\sqrt{4x+5}+3}{4}$
Determine the limit:
$L=\dfrac{\sqrt{4(1)+5}+3}{4}=\dfrac{\sqrt 9+3}{4}=\dfrac{6}{4}=\dfrac{3}{2}$
