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  • Without using the method of L' Hospital, find the limit. <img alt="\lim _{x \rightarrow 0} \frac{(\sqrt{e})^{f(x)}-1}{f(x)}, \text { where } f(x)=\sin x-\tan x" src="https://entrancec

Without using the method of L' Hospital, find the limit.

\lim _{x \rightarrow 0} \frac{(\sqrt{e})^{f(x)}-1}{f(x)}, \text { where } f(x)=\sin x-\tan x

Option: 1

\pm \frac{1}{2}


Option: 2

\frac{1}{2}


Option: 3

-\frac{1}{2}


Option: 4

Cannot be determined


Answers (1)

best_answer

The value of \lim _{x \rightarrow 0} f(x) is

$$ \begin{aligned} & \lim _{x \rightarrow 0} f(x) \\ & =\lim _{x \rightarrow 0}(\sin x-\tan x) \\ & =\sin 0-\tan 0 \\ & =0-0 \\ & =0 \end{aligned}

Note the following exponential limit.

\cdot \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1

\cdot \: \: \frac{\log _{e}(1+x)}{x}=1 \\ \cdot \: \: \lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\log _{e} a=\ln a

Now, evaluate the following limit using the above formula.

\begin{aligned} & \lim _{x \rightarrow 0} \frac{(\sqrt{e})^{f(x)}-1}{f(x)} \\ & =\lim _{x \rightarrow 0} \frac{e^{\frac{f(x)}{2}}-1}{f(x)} \\ & =\frac{1}{2} \times \lim _{x \rightarrow 0} \frac{e^{\frac{f(x)}{2}}-1}{\frac{f(x)}{2}} \\ & =\frac{1}{2} \times 1 \quad\left[ \lim _{x \rightarrow 0} f(x)=0\right] \\ & =\frac{1}{2} \end{aligned}

Posted by

Kuldeep Maurya

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