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 _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crigh

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}

Option: 1

\frac{1}{2}


Option: 2

-\frac{1}{2}


Option: 3

0


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{\sqrt{1+x}-1}{x}=\frac{0}{0}

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

\frac{d}{d x} x^{n}=n x^{n-1}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\sqrt{1+x}-1)}{\frac{d}{d x}(x)} \\ & =\lim _{x \rightarrow 0} \frac{2 \sqrt{1+x}}{1} \quad\left[\not \frac{d}{d x} x^{n}=n x^{n-1}\right] \\ & \left.=\frac{1}{2 \sqrt{1+0}}\right] \\ & =\frac{1}{2} \end{aligned}

 

Posted by

Ajit Kumar Dubey

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

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question