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  • Using the method of L' Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\left(a^{2}\right)^{3 x}-e^{4 x}}{x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cl

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 0} \frac{\left(a^{2}\right)^{3 x}-e^{4 x}}{x}

Option: 1

\log _{e} a


Option: 2

2\left(3 \log _{e} a\right)


Option: 3

2\left(3 \log _{e} a-2\right)


Option: 4

 Cannot be determined


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{\left(a^{2}\right)^{3 x}-e^{4 x}}{x}=\frac{0}{0}

Here use the L' Hospital rule and also refer to the following formula.

\begin{aligned} & \frac{d}{d x}\left(e^{a x}\right)=a e^{a x} \\ & \frac{d}{d x}\left(b^{x}\right)=b^{x} \ln b \end{aligned}

Now, derive the following: 

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(a^{2}\right)^{3 x}-e^{4 x}}{x} \\ & =\lim _{x \rightarrow 0} \frac{a^{6 x}-e^{4 x}}{x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d x}{d x}\left(a^{6 x}-e^{4 x}\right)}{\frac{d}{d x}(x)} \\ \end{aligned}\begin{aligned} & =\lim _{x \rightarrow 0} \frac{6 a^{6 x} \ln a-4 e^{4 x}}{1} \quad\left[ \frac{d}{d x}\left(e^{a x}\right)=a e^{a x}, \frac{d}{d x}\left(b^{x}\right)=b^{x} \ln b\right] \\ & =6 \times 1 \times \ln a-4 \times 1] \\ & =6 \ln a-4 \quad\left[ \ln a=\log _{e} a\right] \\ & =2(3 \ln a-2) \\ & =2(3 \log a-2) \end{aligned}

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shivangi.shekhar

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