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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{\theta \rightarrow \infty}\left(\theta \log _e\left(1+\frac{3^3}{\theta}\right)\right)" src

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

\lim _{\theta \rightarrow \infty}\left(\theta \log _e\left(1+\frac{3^3}{\theta}\right)\right)

Option: 1

9


Option: 2

\frac{\ln2 }{9}+1


Option: 3

27


Option: 4

3


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.

Assume  \frac{1}{\theta}=x

So, as \theta \rightarrow \infty , x\rightarrow 0 , and the provided limit is

\lim _{\theta \rightarrow \infty}\left(\theta \log _e\left(1+\frac{3^3}{\theta}\right)\right)

\begin{aligned} & =\lim _{x \rightarrow 0}\left(\frac{1}{x} \log _e\left(1+3^3 x\right)\right) \\ & =\lim _{x \rightarrow 0} \frac{\log _e(1+27 x)}{x} \\ & =\frac{0}{0} \end{aligned}

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

\begin{aligned} & \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{\log _e(1+27 x)}{x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(\log _e(1+27 x)\right)}{\frac{d}{d x}(x)} \end{aligned}

\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{27}{(1+27 x)}}{1} \quad\left[0 \frac{d}{d x}\left(\log _8 x\right)=\frac{1}{x}\right] \\ & =\frac{27}{1+27 \times 0} \\ & =27 \end{aligned}

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