Question
Let $$f\left( x \right) = g\left( x \right).\frac{{{e^{\frac{1}{x}}} - {e^{ - \frac{1}{x}}}}}{{{e^{\frac{1}{x}}} + {e^{ - \frac{1}{x}}}}},$$ where $$g$$ is a continuous function then $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$ does not exist if :
A.
$$g\left( x \right)$$ is any constant function
B.
$$g\left( x \right) = x$$
C.
$$g\left( x \right) = {x^2}$$
D.
$$g\left( x \right) = x\,h\left( x \right),$$ where $$h\left( x \right)$$ is a polynomial
Answer :
$$g\left( x \right)$$ is any constant function
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^{\frac{1}{x}}} - {e^{ - \frac{1}{x}}}}}{{{e^{\frac{1}{x}}} + {e^{ - \frac{1}{x}}}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1 - {e^{ - \frac{2}{x}}}}}{{1 + {e^{ - \frac{2}{x}}}}} = 1 \cr
& {\text{and }}\mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{\frac{1}{x}}} - {e^{ - \frac{1}{x}}}}}{{{e^{\frac{1}{x}}} + {e^{ - \frac{1}{x}}}}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{\frac{2}{x}}} - 1}}{{{e^{\frac{2}{x}}} + 1}} = - 1 \cr} $$
Hence, $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$ exists if $$\mathop {\lim }\limits_{x \to 0} g\left( x \right) = 0$$
If $$g\left( x \right) = a \ne 0$$ (constant) then $$\mathop {\lim }\limits_{x \to 0 + } f\left( x \right) = a$$ and $$\mathop {\lim }\limits_{x \to 0 - } f\left( x \right) = - a$$
Thus $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$ doesn't exist in this case.