Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Which of the following limit exists finitely?Option: 1 <img alt="\lim _{x \rightarrow 0^{+}}(x)^{\log _e x}" src="https://entrancecorner.oncodecogs.c

Which of the following limit exists finitely?

Option: 1

\lim _{x \rightarrow 0^{+}}(x)^{\log _e x}


Option: 2

\lim _{x \rightarrow 3} \frac{x^2-9-\sqrt{x^2-6 x+9}}{|x-1|-2}


Option: 3

\lim _{x \rightarrow 1^{+}}([x])^{\frac{1}{x-1}}=  (where [.] denotes the greatest integer function)


Option: 4

 none of these


Answers (1)

best_answer

(a)    \lim _{x \rightarrow 0^{+}}(x)^{\log _e x}=\left(0^{+}\right)^{-\infty}=\infty
        So, limit does not exist finitely.

(b)
        \begin{aligned} & \lim _{x \rightarrow 3} \frac{x^2-9-\sqrt{x^2-6 x+9}}{|x-1|-2} \\ & =\lim _{x \rightarrow 3} \frac{x^2-9-|x-3|}{x-3} \end{aligned}
       Now,                 f\left(3^{+}\right)=\lim _{x \rightarrow 3^{+}} \frac{\left(x^2-9\right)-(x-3)}{(x-3)}=\lim _{x \rightarrow 3^{+}}((x+3)-1)=5
        Also,      f\left(3^{-}\right)=\lim _{x \rightarrow 3^{-}} \frac{\left(x^2-9\right)+(x-3)}{(x-3)}=\lim _{x \rightarrow 3^{-}}((x+3)+1)=7
       So, limit does not exist.


(c)

    \lim _{x \rightarrow 1^{+}}([x])^{\frac{1}{x-1}}=\lim _{x \rightarrow 1^{+}} \frac{1}{x-1}=1

Posted by

Rishi

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

How to crack JEE Mains 2025 in 1 month? Follow study plan, revise high-score topics, experts advice
anam-khan

IIT JEE aspirant from Bihar found hanging in Kota hostel; 17th suicide this year
anam-khan

JEE Main 2025 just a month away; section-wise preparation tips, important topics, strategies

Latest Question