Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow%20%5Cinfty%7D%5Cleft%28x-%5Csqrt%7Bx%5E2&p

\lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right)

Option: 1

\frac{-1}{4}


Option: 2

\frac{-1}{2}


Option: 3

\frac{-1}{3}


Option: 4

\frac{-3}{4}


Answers (1)

best_answer

By multiplying the numerator and denominator by the conjugate of the equation within the square root, you may make the expression simpler:

\begin{aligned} & \lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right) \\ & =\lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right) \cdot \frac{x+\sqrt{x^2+x}}{x+\sqrt{x^2+x}} \end{aligned}

use the identity  (a-b)(a+b)=a^2-b^2 to simplify the numerator:

\begin{aligned} &\lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right) \cdot \frac{x+\sqrt{x^2+x}}{x+\sqrt{x^2+x}}\\ &\lim _{x \rightarrow \infty} \frac{x^2-\left(x^2+x\right)}{x+\sqrt{x^2+x}}=\lim _{x \rightarrow \infty} \frac{-1}{1+\sqrt{1+\frac{1}{x}}} \end{aligned}

\begin{aligned} & =\frac{-1}{1+\sqrt{1+\frac{1}{\infty}}} \\ & =\frac{-1}{1+\sqrt{1+0}}=\frac{-1}{1+1} \\ & =\frac{-1}{2} \end{aligned}

 

 

Posted by

jitender.kumar

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