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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{(\sqrt[4]{16})^x+\log _e(1+2 x)-\left(e^2\right)^x}{x}" src="https:/

Using the method of L’ Hospital, evaluate the limit.

\lim _{x \rightarrow 0} \frac{(\sqrt[4]{16})^x+\log _e(1+2 x)-\left(e^2\right)^x}{x}

 

Option: 1

In 2


Option: 2

\frac{\ln 2}{4}+1

 


Option: 3

-2 \ln 2


Option: 4

- \ln 2


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}, \text { 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{(\sqrt[4]{16})^x+\log _e(1+2 x)-\left(e^2\right)^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 \\ & \frac{d}{d x}\left(\log _e x\right)=\frac{1}{x} \end{aligned}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{(\sqrt[4]{16})^x+\log _e(1+2 x)-\left(e^2\right)^x}{x} \\ & =\lim _{x \rightarrow 0} \frac{16^{\frac{x}{4}}+\log _e(1+2 x)-e^{2 x}}{x} \\ \end{aligned}

\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(16^4+\log _0(1+2 x)-e^{2 x}\right)}{\frac{d}{d x}(x)} \\ & =\lim _{x \rightarrow 0} \frac{\frac{1}{4} 16^{\frac{x}{4}} \ln 16+\frac{2}{(1+2 x)}-2 e^{2 x}}{1}\left[0 \frac{d}{d x}\left(b^x\right)=b^x \ln b, \frac{d}{d x}\left(\log _e x\right)=\frac{1}{x}, \frac{d}{d x}\left(e^{a x}\right)=a e^{a x}\right] \end{aligned}

\begin{aligned} & =\frac{1}{4} \times 1 \times \ln 16+2-2 \times 1 \\ & =\frac{1}{4} \ln 2^4 \\ & =\ln 2 \end{aligned}

 

Posted by

HARSH KANKARIA

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