Question

Let $f\left(x\right)=g\left(x\right)\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}$ and $x\ne 0$ where g is a continuous function.

Then ${lim}_{x\to 0}f\left(x\right)$ exists if?

• A. $g\left(x\right)$ is any polynomial
• B. $g\left(x\right)=x+4$
• C. $g\left(x\right)=x^2$
• D. $g\left(x\right)=2+3x+4x^2$

Answer 1

The correct option is C $g\left(x\right)=x^2$

$f\left(x\right)=g\left(x\right)\times \frac{e^{2/x}-1}{e^{2/x}+1}$

$=g\left(x\right)\times (1-\frac{2}{e^{2/x}+1})$

$=g\left(x\right)-\frac{2g\left(x\right)}{e^{2/x}+1}$

At $x=0$

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}(g\left(x\right)-\frac{2g\left(x\right)}{e^{2/x}+1})$

$=g\left(0\right)-\frac{2g\left(0\right)}{1/\infty +1}$ $as(x\to 0^{-};\frac{1}{x}\to -\infty )$

$=g\left(0\right)-2g\left(0\right)=-g\left(0\right)$

${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}(g\left(x\right)-\frac{2g\left(x\right)}{e^{2/x}+1})$

$=g\left(0\right)-\frac{2g\left(0\right)}{e^{\infty }+1}$ $as(x\to 0^{+};\frac{1}{x}\to \infty )$

$=g\left(0\right)-0=g\left(0\right)$

For $f\left(x\right)$ to be continuous at $x=0$

LHL$=$RHL

$\Rightarrow -g\left(0\right)=g\left(0\right)\Rightarrow g\left(0\right)=0$

Only option C satisfies