# Find the following limits or state that they do not exist. \lim_{x\to1^+}

BenoguigoliB 2021-11-05 Answered
Find the following limits or state that they do not exist.
$\underset{x\to {1}^{+}}{lim}\frac{x-1}{\sqrt{{x}^{2}-1}}$
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## Expert Answer

Obiajulu
Answered 2021-11-06 Author has 98 answers
Given: $\underset{x\to {1}^{+}}{lim}\frac{x-1}{\sqrt{{x}^{2}-1}}$
for evaluating given limit we substitute x=1+h, where $h\to 0$
so,
$\underset{x\to {1}^{+}}{lim}\frac{x-1}{\sqrt{{x}^{2}-1}}=\underset{h\to {0}^{+}}{lim}\frac{1+h-1}{\sqrt{{\left(1+h\right)}^{2}-1}}$
$=\underset{h\to {0}^{+}}{lim}\frac{h}{\sqrt{{1}^{2}+2\left(1\right)\left(h\right)+{h}^{2}-1}}$
$=\underset{h\to {0}^{+}}{lim}\frac{h}{\sqrt{1+2h+{h}^{2}-1}}$
$=\underset{h\to {0}^{+}}{lim}\frac{h}{\sqrt{{h}^{2}+2h}}$
$=\underset{h\to {0}^{+}}{lim}\frac{h}{h\sqrt{1+\frac{2}{h}}}$
$=\underset{h\to {0}^{+}}{lim}\frac{1}{\sqrt{1+\frac{2}{h}}}$
$=\frac{1}{\sqrt{1+\mathrm{\infty }}}$
$=0$
hence, given limit is equal to 0.
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