# Evaluate the following limits \lim_{x\to\infty}(\sqrt{x^2+4x+1}-x)

Evaluate the following limits
$\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt{{x}^{2}+4x+1}-x\right)$
$\underset{x\to -\mathrm{\infty }}{lim}\left(\sqrt{{x}^{2}+4x+1}-x\right)$
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We have to evaluate the limit
a) $\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt{{x}^{2}+4x+1}-x\right)$
$=\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{{x}^{2}+4x+1}-x}{\sqrt{{x}^{2}+4x+1}+x}×\sqrt{{x}^{2}+4x+1}+x$
$=\underset{x\to \mathrm{\infty }}{lim}\frac{x\left(4+\frac{1}{x}\right)}{x\left(\sqrt{1+\frac{4}{x}+\frac{1}{{x}^{2}}}+1\right)}$
$-\frac{4}{\sqrt{1}+1}=\frac{4}{2}=\frac{1}{2}$
b) $\underset{x\to -\mathrm{\infty }}{lim}\left(\sqrt{{x}^{2}+4x+1}-x\right)$
pu x=-x
$⇒\underset{x\to -\mathrm{\infty }}{lim}\left(\sqrt{{x}^{2}+4x+1}-x\right)=\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt{{x}^{2}-4x+1}+x\right)$
$=\underset{x\to \mathrm{\infty }}{lim}\sqrt{{x}^{2}-4x+1}+\underset{x\to \mathrm{\infty }}{lim}x$
$\underset{x\to \mathrm{\infty }}{lim}\sqrt{{x}^{2}\left(1-\frac{4}{x}+\frac{1}{{x}^{2}}\right\}+\mathrm{\infty }}$
$=\mathrm{\infty }+\mathrm{\infty }$
$\therefore \underset{x\to -\mathrm{\infty }}{lim}\left(\sqrt{{x}^{2}+4x+1}-x\right)=\mathrm{\infty }$