Question

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

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

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

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

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 $\underset{x\to 0}{lim}f\left(x\right)$ will exist if $g\left(x\right)=x^2$ or $g\left(x\right)=x^{3}h\left(x\right)$ is a polynomial.