Arectemieryf0

2022-07-27

$lim\left(\sqrt{{x}^{2}+1}-\sqrt{2}\right)/\left(x-1\right)$
x approaches 1

jbacapzh

$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=$
Evaluating gives 0/0, indicating a common factor of x-1. So we need to manipulate this a little to exposethis, so we can cancel it and then try to evaluate the limit. Multiplying the top and bottom by the conjugate
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\underset{x\to 0}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}\ast \frac{\sqrt{{x}^{2}+1}+\sqrt{2}}{\sqrt{{x}^{2}+2}+\sqrt{2}}$
Simplifying
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\underset{x\to 0}{lim}\frac{{x}^{2}+1-2}{\left(x-1\right)\left[\sqrt{{x}^{2}+1}+\sqrt{2}\right]}$
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\underset{x\to 0}{lim}\frac{{x}^{2}-1}{\left(x-1\right)\left[\sqrt{{x}^{2}+1}+\sqrt{2}\right]}$
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\underset{x\to 0}{lim}\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left[\sqrt{{x}^{2}+1}+\sqrt{2}\right]}$
And there it is, canceling
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\underset{x\to 0}{lim}\frac{x+1}{\sqrt{{x}^{2}+1}+\sqrt{2}}$
Evaluating
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\frac{1+1}{\sqrt{1+1}+\sqrt{2}}$
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\frac{2}{\sqrt{2}+\sqrt{2}}$
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\frac{2}{2+\sqrt{2}}$
$\underset{x\to 1}{lim}\frac{\sqrt{{x}^{2}+1}-\sqrt{2}}{x-1}=\frac{1}{\sqrt{2}}$

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