Find the following limits or state that they do not exist.

$\underset{x\to {1}^{+}}{lim}\frac{x-1}{\sqrt{{x}^{2}-1}}$

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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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,

$\underset{x\to {1}^{+}}{lim}\frac{x-1}{\sqrt{{x}^{2}-1}}=\underset{h\to {0}^{+}}{lim}\frac{1+h-1}{\sqrt{{(1+h)}^{2}-1}}$

$=\underset{h\to {0}^{+}}{lim}\frac{h}{\sqrt{{1}^{2}+2\left(1\right)\left(h\right)+{h}^{2}-1}}$

$=\underset{h\to {0}^{+}}{lim}\frac{h}{\sqrt{1+2h+{h}^{2}-1}}$

$=\underset{h\to {0}^{+}}{lim}\frac{h}{\sqrt{{h}^{2}+2h}}$

$=\underset{h\to {0}^{+}}{lim}\frac{h}{h\sqrt{1+\frac{2}{h}}}$

$=\underset{h\to {0}^{+}}{lim}\frac{1}{\sqrt{1+\frac{2}{h}}}$

$=\frac{1}{\sqrt{1+\mathrm{\infty}}}$

$=0$

hence, given limit is equal to 0.

for evaluating given limit we substitute x=1+h, where

so,

hence, given limit is equal to 0.

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