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Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 0} \frac{\sin \pi x}{x \sqrt{1+x}}

Option: 1

\frac{2}{3}


Option: 2

\pm \frac{2}{3}


Option: 3

\pi


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

\lim _{x \rightarrow 0} \frac{\sin \pi x}{x \sqrt{1+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} \\ & \frac{d}{d x}(u v)=\frac{dv}{d x} u+\frac{du}{d x} v \end{aligned}

Now, derive the following:

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

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Gaurav

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