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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 Marks
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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