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  • Using the method of L' Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\sin 2 x}{\sqrt{1+x}-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 0} \frac{\sin 2 x}{\sqrt{1+x}-1}

Option: 1

1


Option: 2

2


Option: 3

4


Option: 4

0


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 2 x}{\sqrt{1+x}-1}=\frac{0}{0}

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

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

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin 2 x}{\sqrt{1+x}-1} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x} \sin 2 x}{\frac{d}{d x}(\sqrt{1+x}-1)} \\ & =\lim _{x \rightarrow 0} \frac{2 \cos 2 x}{\frac{1}{2 \sqrt{1+x}}} \quad\left[ \frac{d}{d x} x^{n}=n x^{n-1}, \frac{d}{d x}(\sin x)=\cos x\right] \\ & =\frac{2 \times 1}{\frac{1}{2 \sqrt{1+0}}} \\ & =4 \end{aligned}

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avinash.dongre

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