Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{\theta \rightarrow x}\left(\frac{1}{\theta \log _e\left(\frac{\theta+3}{\theta}\right)}+1\r

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

\lim _{\theta \rightarrow x}\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

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

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

=\frac{1}{\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(\log _x(1+3 x)\right)}{d}(x)}+1

\begin{aligned} & =\frac{1}{\lim _{x \rightarrow 0} \frac{3}{1+3 x}}+1 \\ & =\frac{1}{3}+1 \\ & =1 \frac{1}{3} \\ & \end{aligned}

Posted by

shivangi.shekhar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?
anam-khan

JEE Main 2025 session 1 official answer key out; challenge by February 6; how to calculate NTA score?
anam-khan

JEE Main 2025: NIT Raipur's CSE cut-offs drop; closing ranks in popular branches

Latest Question