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 $\displaystyle \lim_{x \to 0} f (x)$ does not exist if

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

Answer 1

The Correct Option is A

$\displaystyle \lim_{x \to 0^+}$ $\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}=\displaystyle \lim_{x \to 0^+}$$\frac{1-e^{-2/x}}{1+e^{-2/x}}=1$ and $\displaystyle \lim_{x \to 0^-}$$\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}=\displaystyle \lim_{x \to 0^-}$$\frac{e^{2/x}-1}{e^{2/x}+1}=-1.$ Hence $\displaystyle \lim_{x \to 0} f (x)$ exists if $\displaystyle \lim_{x \to 0}g(x)=0.$ If $g(x) = a \ne 0$ (constant) then $\displaystyle \lim_{x \to 0^+}f (x)=a$ and $\displaystyle \lim_{x \to 0^-}f (x)=-a.$ Thus $\displaystyle \lim_{x \to 0}f(x)$ doesn?? exist in this case. $\therefore \displaystyle \lim_{x \to 0}f(x)$ exists in case of (b), (c) and (d) each.