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  • Without using the method of L' Hospital, find the limit.   <img alt="\lim _{\theta \rightarrow \infty}\left(\frac{1}{\theta \log _{e}\left(\frac{\theta+3}{\theta}\right)}+1\

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

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

Option: 1

9


Option: 2

\frac{1}{3}


Option: 3

1\frac{1}{3}


Option: 4

3


Answers (1)

best_answer

Note the following exponential limit.

\lim _{x \rightarrow 0}\frac{\log _{e}(1+x)}{x}=1

Assume   \frac{1}{\theta}=x

So, as   \theta \rightarrow \infty, x \rightarrow 0

Now, evaluate the following limit using the above formula and the assumption.  

\begin{aligned} & \lim _{\theta \rightarrow \infty}\left(\frac{1}{\theta \log _{e}\left(\frac{\theta+3}{\theta}\right)}+1\right) \\ & =\frac{1}{\lim _{\theta \rightarrow \infty}\left(\theta \log _{e}\left(1+\frac{3}{\theta}\right)\right)}+1 \\ & =\frac{1}{\lim _{x \rightarrow 0} \frac{1}{x} \log _{e}(1+3 x)}+1 \\ \end{aligned}\begin{aligned} & =\frac{1}{3 \times \lim _{x \rightarrow 0} \frac{\log _{e}(1+3 x)}{3 x}}+1 \\ & =\frac{1}{3 \times 1}+1 \\ & =\frac{4}{3} \\ & =1 \frac{1}{3} \end{aligned}

Posted by

manish painkra

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