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Using the method of L' Hospital, evaluate the limit.  \lim _{x \rightarrow 0} \frac{\sqrt{3+2 x}-\sqrt{3-2 x}}{\sin (\pi x)}

Option: 1

\frac{2}{3}


Option: 2

\pm \frac{2}{3}


Option: 3

\frac{1}{3}


Option: 4

\frac{2}{\sqrt{3} \pi}


Answers (1)

best_answer

Apply the L' Hospital's Rule, when \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, etc. [an intermediate form] in the following way

- Differentiate \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, provided that the limit \lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} exists.

- Otherwise, go on differentiating till the limit with the determinate form is achieved.

The provided limit is

\frac{\sqrt{3+2 x}-\sqrt{3-2 x}}{\sin (\pi x)}=\frac{0}{0}

Here use the L' Hospital rule and also refer to the following formula. 

\begin{aligned} & \frac{d}{d x}(\sin x)=\cos x \\ & \frac{d}{d x} x^{n}=n x^{n-1} \end{aligned}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sqrt{3+2 x}-\sqrt{3-2 x}}{\sin (\pi x)} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\sqrt{3+2 x}-\sqrt{3-2 x})}{\frac{d}{d x}(\sin \pi x)} \\ & =\lim _{x \rightarrow 0} \frac{\frac{1 \times 2}{2 \sqrt{3+2 x}}-\frac{1 \times(-2)}{2 \sqrt{3-2 x}}}{\pi \cos \pi x} \quad\left[ \frac{d}{d x} x^{n}=n x^{n-1}, \frac{d}{d x}(\sin x)=\cos x\right] \\ \end{aligned}\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{3+2 x}}+\frac{1}{\sqrt{3-2 x}}}{\pi \cos \pi x} \\ & =\frac{\frac{1}{\sqrt{3+0}}+\frac{1}{\sqrt{3-0}}}{\pi \times 1} \\ & =\frac{2}{\sqrt{3} \pi} \end{aligned}

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Rishabh

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