Question

Assertion(A):
$f\left(x\right)=x(\frac{1+e^{1/x}}{1-e^{1/x}}\left)\right(x\ne 0)$ , $f\left(0\right)=0$ is continuous at $x=0$.
Reason(R) A function is said to be continuous at $a$ if both limits are exists and equal to $f\left(a\right)$ .

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true and R is not the correct explanation of A
• C. A is true but R is false
• D. R is true but A is false

Answer 1

The correct option is A Both A and R are true and R is the correct explanation of A

Assertion : $f\left(x\right)=x(\frac{1+e^{1/x}}{1-e^{1/x}}\left)\right(x\ne 0)$
$LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}x\left(\frac{1+e^{1/x}}{1-e^{1/x}}\right)={lim}_{h\to 0}(-h)\left(\frac{1+e^{-1/h}}{1-e^{-1/h}}\right)=0$
$RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}x\left(\frac{1+e^{1/x}}{1-e^{1/x}}\right)$

$={lim}_{h\to 0}h\left(\frac{1+e^{1/h}}{1-e^{1/h}}\right)={lim}_{h\to 0}h\left(\frac{e^{-1/h}+1}{e^{-1/h}-1}\right)=0$
Given $f\left(0\right)=0$
So, $LHL=RHL=f\left(0\right)$
Hence, $f\left(x\right)$ is continuous at $x=0$
Also, reason is true.