Answer
$\lim_{x\to1}\dfrac{x^{2}-1}{\sqrt{x}-1}=4$
The graph is shown below:
Work Step by Step
$\lim_{x\to1}\dfrac{x^{2}-1}{\sqrt{x}-1}$
Try to evaluate the limit applying direct substitution:
$\lim_{x\to1}\dfrac{x^{2}-1}{\sqrt{x}-1}=\dfrac{1^{2}-1}{\sqrt{1}-1}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Factor the numerator and simplify:
$\lim_{x\to1}\dfrac{x^{2}-1}{\sqrt{x}-1}=\lim_{x\to1}\dfrac{(x-1)(x+1)}{\sqrt{x}-1}=...$
$...=\lim_{x\to1}\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)(x+1)}{\sqrt{x}-1}=...$
$...=\lim_{x\to1}(\sqrt{x}+1)(x+1)=...$
Try to evaluate the limit applying direct substitution again:
$\lim_{x\to1}(\sqrt{x}+1)(x+1)=(\sqrt{1}+1)(1+1)=4$
