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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limits:
$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+4 x+16}-\sqrt{x^2+16}\right]$
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Answer

$\lim _{x \rightarrow \infty} \sqrt{x^2+4 x+16}-\sqrt{x^2+16}$
$=\lim _{x \rightarrow \infty} \frac{\left(\sqrt{x^2+4 x+16}-\sqrt{x^2+16}\right)\left(\sqrt{x^2+4 x+16}+\sqrt{x^2+16}\right)}{\left(\sqrt{x^2+4 x+16}+\sqrt{x^2+16}\right)} $
...[By rationalization]
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x}{x}}{\sqrt{\frac{x^2+4 x+16}{x^2}+\sqrt{\frac{x^2+16}{x^2}}}}$
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x}{x}}{\sqrt{\frac{x^2+4 x+16}{x^2}}+\sqrt{\frac{x^2+16}{x^2}}}$
$=\lim _{x \rightarrow \infty} \frac{4}{\sqrt{1+\frac{4}{x}+\frac{16}{x^2}}+\sqrt{1+\frac{16}{x^2}}}$
$=\frac{4}{\lim _{x \rightarrow \infty} \sqrt{1+\frac{4}{x}+\frac{16}{x^2}}+\lim _{x \rightarrow \infty} \sqrt{1+\frac{16}{x^2}}}$
$=\frac{4}{\sqrt{1+0+0}+\sqrt{1+0}} \quad \cdots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, k>0\right]$
$=\frac{4}{2}$
$=2$

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