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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{4\text{x}^2-7\text{x}}+2\text{x}\big)$
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Answer

$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{4\text{x}^2-7\text{x}}+2\text{x}\big)$
Substitute y = -x
$=\lim\limits_{\text{y}\rightarrow\infty}\Big(\sqrt{4\text{y}^2+7\text{y}}-2\text{y}\Big)$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{\Big(\sqrt{4\text{y}^2+7\text{y}}-2\text{y}\Big)\Big(\sqrt{4\text{y}^2+7\text{y}}+2\text{y}\Big)}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{\Big(4\text{y}^2+7\text{y}+4\text{y}^2\big)}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{(7\text{y})}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{7}{\sqrt{4+\frac{7}{\text{y}}}+2}$
$=\frac{7}{2+2}=\frac{7}{4}$

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