Evaluate the following limit: _ x - ( x^2-8x +x ) - Vidyadip

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{\text{x}^2-8\text{x}}+\text{x}\big)$

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Answer

$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{\text{x}^2-8\text{x}}+\text{x}\big)$
$=\lim\limits_{\text{y}\rightarrow\infty}\Big(\sqrt{\text{y}^2+8\text{y}}-\text{y}\Big),$ where y = -x on rationalising
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{\big(\sqrt{\text{y}^2+8\text{y}}-\text{y}\big)\big(\sqrt{\text{y}^2+8\text{y}}+\text{y}\big)}{\big(\sqrt{\text{y}^2+8\text{y}}+\text{y}\big)}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{\text{y}^2+8\text{y}+\text{y}^2}{\sqrt{\text{y}^2+8\text{y}}+\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{8\text{y}}{\text{y}\sqrt{1+\frac{8}{\text{y}}}+\text{y}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{8}{\sqrt{1+\frac{8}{\text{y}}}+1}$
$=\frac{8}{1+1}=\frac{8}{2}$
$=4$

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