If $L$ $=\lim _{x \rightarrow 0} \frac{a-\sqrt{a^{2}-x^{2}}-\frac{x^{2}}{4}}{x^{4}}, a0 .$ If $L$ is finite,then

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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

Aicharya · Accepted Answer

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 $\Rightarrow \frac{1}{a}-\frac{1}{2}=0 \Rightarrow a=2$

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…

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