Question

Let $f\left(x\right)=g\left(x\right).\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}},$ where g is a continuous function then $\underset{x\to 0}{\lim }f(x)$ does not exist if

• A. $g(x)$ is any constant function
• B. $g(x)=x$
• C. $g(x)=x^2$
• D. $g(x)=xh(x)$, where $h(x)$ is a polynomial

Answer 1

Answer: (A)

$\underset{x\to 0^{+}}{\lim }$ $\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}=\underset{x\to 0^{+}}{\lim }$$\frac{1-e^{-2/x}}{1+e^{-2/x}}=1$
and $\underset{x\to 0^{-}}{\lim }$$\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}=\underset{x\to 0^{-}}{\lim }$$\frac{e^{2/x}-1}{e^{2/x}+1}=-1.$
Hence $\underset{x\to 0}{\lim }f(x)$ exists if $\underset{x\to 0}{\lim }g(x)=0.$
If $g(x)=a\ne 0$ (constant) then
$\underset{x\to 0^{+}}{\lim }f(x)=a$ and $\underset{x\to 0^{-}}{\lim }f(x)=-a.$
Thus $\underset{x\to 0}{\lim }f(x)$ doesn’t exist in this case.
$\therefore \underset{x\to 0}{\lim }f(x)$ exists in case of (b), (c) and (d) each.