Question

Let $f\left(x\right)=g^{\prime }\left(x\right)\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}$, where $g^{\prime }$ is the derivative of $g$ and is a continuous function, then $\underset{x\to 0}{lim}f\left(x\right)$ exists if

• A. $g\left(x\right)$ is a polynomial
• B. $g\left(x\right)=x$
• C. $g\left(x\right)=x^2$
• D. $g\left(x\right)=x^{3}h\left(x\right)$, where $h\left(x\right)$ is a polynomial

Answer 1

Answer: (C), (D)

The correct options are
C $g\left(x\right)=x^2$
D $g\left(x\right)=x^{3}h\left(x\right)$, where $h\left(x\right)$ is a polynomial

Given, $f\left(x\right)=g^{\prime }\left(x\right)=$ $\underset{x\to 0^{+}}{lim}\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{1}{x}}}=\underset{x\to 0^{+}}{lim}\frac{1-e^{\frac{-2}{x}}}{1+e^{\frac{-2}{x}}}=1$

and, $\underset{x\to 0^{-}}{lim}\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}=\underset{x\to 0^{-}}{lim}\frac{e^{\frac{2}{x}}-1}{e^{\frac{2}{x}}+1}=-1$

Hence, $\underset{x\to 0}{lim}f\left(x\right)$ exists if $g^{\prime }\left(0\right)=0$

If $g\left(x\right)=ax+b,a\ne 0$ then $\underset{x\to 0^{+}}{lim}f\left(x\right)=a$ and $\underset{x\to 0^{-}}{lim}f\left(x\right)=-a.$

Hence, $\underset{x\to 0}{lim}f\left(x\right)$ exists if $g\left(x\right)=x^2$

or, $g\left(x\right)=x^{3}h\left(x\right),$ where $h\left(x\right)$ is a polynomial.