Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Limits of functions having built-in limit with them If <img alt="\mathrm{f(x)=\lim _{n \rightarrow \infty} \frac{\sin \left(\pi x^2\right)+(x+2)^n \cdot \tan x}{x^2+(x+2)^n}}" src="https://e

Limits of functions having built-in limit with them
If \mathrm{f(x)=\lim _{n \rightarrow \infty} \frac{\sin \left(\pi x^2\right)+(x+2)^n \cdot \tan x}{x^2+(x+2)^n}}, then find \mathrm{\lim _{x \rightarrow-1} f(x)} (where \mathrm{n \in N} )
 

Option: 1

0


Option: 2

\frac{2}{3}


Option: 3

\infty


Option: 4

none of these


Answers (1)

best_answer

This can be termed as limit of a function \mathrm{f(x)} having built-in limit.
Now, \mathrm{ f(x)=\lim _{n \rightarrow \infty} \frac{\sin \left(\pi x^2\right)+(x+2)^n \cdot \tan x}{x^2+(x+2)^n} }

Clearly the part affected is \mathrm{ (x+2)^n }
Now,
when \mathrm{ \lim _{n \rightarrow \infty}(x+2)^n=\left\{\begin{array}{ccc} \infty & x+2>1 \Rightarrow & x>-1 \\ 1 & x+2=1 \Rightarrow & x=-1 \\ 0 & 0<x+2<1 \Rightarrow & -2<x<-1 \end{array}\right. }

\mathrm{ f(x)= \begin{cases}\lim _{n \rightarrow \infty} \frac{\frac{\sin \left(\pi x^2\right)}{(x+2)^n}+\tan x}{x^2} & \text { for } x>-1 \\ \lim _{n \rightarrow \infty} \frac{\sin \pi x^2+0 \tan x}{x^2+0} & \text { for }-2<x<-1\end{cases} }

\mathrm{\therefore \quad f(x) = \begin{cases}\tan x \quad\quad\: x>-1 \: \: \frac{\sin \pi x^2}{x^2}-2<x<-1\end{cases} }

\mathrm{\lim _{x \rightarrow-1^{-}} f(x) =\lim _{x \rightarrow-1^{+}}(\tan x)=-\tan 1 }

\mathrm{\lim _{x \rightarrow-1^{-}} f(x) =\lim _{x \rightarrow-1^{-}} \frac{\sin \left(\pi x^2\right)}{x^2}=0}

Hence option 1 is correct.


 

Posted by

Gunjita

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 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question