Test Index

AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Mathematics

© examsnet.com

Question : 59 of 160

Marks: +1, -0

If

\lim\limits_{x\to0^{+}}x^{2}\left(\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right)=k

\lim\limits_{x\to0^{-}}x^{2}\left(\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right)=1

, then

$k=l$
$k=1, l=-1$
$k=-1, l=1$
$k\neq l\neq\pm1$

Solution: 👈: Video Solution

k=\lim\limits_{x\to0^{+}}x^{2}\left[\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right]

Let

\frac{1}{x}=t

x\to0^{+}, t\to\infty

k=\lim\limits_{t\to\infty}\frac{1}{t^{2}}\left[\frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}\right]

=\lim\limits_{x\to\infty}\frac{1}{t^{2}}\left[\frac{1-e^{-2t}}{1+e^{-2t}}\right]

=0

\dots\left[\begin{array}{l}\because\lim\limits_{t\to\infty}\;\frac{1}{e^{2t}}=0\\text{and }\lim\limits_{t\to\infty}\;\frac{1}{t^{2}}=0\end{array}\right]

l=\lim\limits_{x\to0^{-}}x^{2}\left[\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right]

Let

\frac{1}{x}=y

x\to0^{-}, y\to-\infty

\therefore l=\lim\limits_{y\to\infty}\frac{1}{y^{2}}\left[\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\right]=0

\dots\text{[same as above]}

\therefore k=l

© examsnet.com

Go to Question:

Prev Question

Next Question