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  • Without using the method of L' Hospital, find the limit. <img alt="\lim _{x \rightarrow 0} \frac{\left(e^{2}\right)^{f(x)}-e^{4 f(x)}}{f(x)}, \text { where } \lim _{x \rightarrow 0} f

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

\lim _{x \rightarrow 0} \frac{\left(e^{2}\right)^{f(x)}-e^{4 f(x)}}{f(x)}, \text { where } \lim _{x \rightarrow 0} f(x)=0

Option: 1

\pm 2


Option: 2

2


Option: 3

-2


Option: 4

 Cannot be determined


Answers (1)

best_answer

Note the following exponential limit.

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

Now, evaluate the following limit using the above formula.

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(e^{2}\right)^{f(x)}-e^{4 f(x)}}{f(x)} \\ & =\lim _{x \rightarrow 0} \frac{e^{2 f(x)}-e^{4 f(x)}-1+1}{f(x)} \\ & =2 \times \lim _{x \rightarrow 0} \frac{e^{2 f(x)}-1}{2 f(x)}-4 \times \lim _{x \rightarrow 0} \frac{e^{4 f(x)}-1}{4 f(x)} \\ & =2 \times 1-4 \times 1\left[\because \lim _{x \rightarrow 0} f(x)=0\right] \\ & =-2 \end{aligned}

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HARSH KANKARIA

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