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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)^{x}-\sqrt{a^{x}}}{x}" src="https://entrancecorner.oncodecogs.com/gif.lat

Without using the method of L' Hospital, find the limit. \lim _{x \rightarrow 0} \frac{\left(e^{2}\right)^{x}-\sqrt{a^{x}}}{x}

Option: 1

2-\log _{e} a


Option: 2

2-\frac{1}{2} \log _{e} a


Option: 3

\frac{1}{2} \log _{e} a


Option: 4

Cannot be determined


Answers (1)

best_answer

Note the following exponential limit.

\begin{aligned} & \text {. } \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1 \\ & \text {. } \lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\log _{e} a \end{aligned}

Now, evaluate the following limit using the above formula. 

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(e^{2}\right)^{x}-\sqrt{a^{x}}}{x} \\ & =\lim _{x \rightarrow 0} \frac{e^{2 x}-a^{\frac{x}{2}}-1+1}{x} \\ \end{aligned}\begin{aligned} & =2 \times \lim _{x \rightarrow 0} \frac{e^{2 x}-1}{2 x}-\frac{1}{2} \times \lim _{x \rightarrow 0} \frac{a^{\frac{x}{2}}-1}{\frac{x}{2}} \\ & =2 \times 1-\frac{1}{2} \times \log _{e} a \\ & =2-\frac{1}{2} \log _{e} a \end{aligned}

 

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Rishabh

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