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If $L$ $$
=\lim _{x \rightarrow 0} \frac{a-\sqrt{a^{2}-x^{2}}-\frac{x^{2}}{4}}{x^{4}}, a>0 .
$$
If $L$ is finite,then
Options:
=\lim _{x \rightarrow 0} \frac{a-\sqrt{a^{2}-x^{2}}-\frac{x^{2}}{4}}{x^{4}}, a>0 .
$$
If $L$ is finite,then
Solution:
1076 Upvotes
Verified Answer
The correct answer is:
$a=2$
Given, $L=\lim _{x \rightarrow 0} \frac{a-\sqrt{a^{2}-x^{2}}-\frac{x^{2}}{4}}{x^{4}}\left[\frac{0}{0}\right.$ form $]$
$\frac{0-\frac{(0-2 x)}{2 \sqrt{a^{2}-x^{2}}}-\frac{2 x}{4}}{4 x^{3}}$ [by using $L^{\prime}$ ' hospital rule]
$\lim _{x \rightarrow 0} \frac{x\left(\frac{1}{\sqrt{a^{2}-x^{2}}}-\frac{1}{2}\right)}{4 x^{3}}=\lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{a^{2}-x^{2}}}-\frac{1}{2}}{4 x^{2}}$
$\left(\frac{\frac{1}{a}-\frac{1}{2}}{0}\right.$ form $)$
For limit to be exist $\Rightarrow \frac{1}{a}-\frac{1}{2}=0 \Rightarrow a=2$
$\frac{0-\frac{(0-2 x)}{2 \sqrt{a^{2}-x^{2}}}-\frac{2 x}{4}}{4 x^{3}}$ [by using $L^{\prime}$ ' hospital rule]
$\lim _{x \rightarrow 0} \frac{x\left(\frac{1}{\sqrt{a^{2}-x^{2}}}-\frac{1}{2}\right)}{4 x^{3}}=\lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{a^{2}-x^{2}}}-\frac{1}{2}}{4 x^{2}}$
$\left(\frac{\frac{1}{a}-\frac{1}{2}}{0}\right.$ form $)$
For limit to be exist $\Rightarrow \frac{1}{a}-\frac{1}{2}=0 \Rightarrow a=2$
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