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

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

Option: 1

\frac{1}{a}


Option: 2

a^{a}


Option: 3

0


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^{1-\sin x}-a^{1-\tan x}}{\sin 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: 

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

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Rishabh

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