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

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{\tan x}

Option: 1

\frac{1}{a}


Option: 2

a^{a}


Option: 3

\ln a


Option: 4

\frac{1}{\ln a^{a}}


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{a^{\sin x}-1}{\tan 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}\left(b^{x}\right)=b^{x} \ln b \\ & \frac{d}{d x}(\tan x)=\sec ^{2} x \end{aligned}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{\tan x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(a^{\sin x}-1\right)}{\frac{d}{d x} \tan x} \\ & =\lim _{x \rightarrow 0}\left(\frac{a^{\sin x} \ln a \times \frac{d}{d x}(\sin x)-0}{\sec ^{2} x}\right) \\ & =\lim _{x \rightarrow 0}\left(\frac{a^{\sin x} \ln a \times \cos x}{\sec x}\right) \\ & =\frac{a^{\sin 0} \ln a \times \cos 0}{\sec ^{2} 0} \\ & =\ln a \end{aligned}

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rishi.raj

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