Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • How can I evaluate the following limit <img alt="\mathrm{\lim _{x \rightarrow \infty} \frac{\sqrt{4+x^{2}}+x}{x^{2} \sin \left(\frac{1}{x}\right)}}" src="https://entrancecorner.oncod

How can I evaluate the following limit

\mathrm{\lim _{x \rightarrow \infty} \frac{\sqrt{4+x^{2}}+x}{x^{2} \sin \left(\frac{1}{x}\right)}}

Option: 1

1


Option: 2

3


Option: 3

1


Option: 4

-1


Answers (1)

best_answer

divide both the numerator and the denominator by \mathrm{x} and you get:

\mathrm{\lim _{x \rightarrow \infty} \frac{\sqrt{4+x^{2}}+x}{x^{2} \sin \left(\frac{1}{x}\right)}=\lim _{x \rightarrow \infty} \frac{\sqrt{\frac{4}{x^{2}}+1}+1}{x \sin \left(\frac{1}{x}\right)}}


\mathrm{\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right)=\text{1 is very known limit and it is obvious that }}
\mathrm{\lim _{x \rightarrow \infty} \sqrt{\frac{4}{x^{2}}+1}+1=\sqrt{1}+1=2 }

hence you have
\mathrm{\lim _{x \rightarrow \infty} \frac{\sqrt{4+x^{2}}+x}{x^{2} \sin \left(\frac{1}{x}\right)}=\frac{2}{1}=2}

Posted by

Suraj Bhandari

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question