Answer
$2$
Work Step by Step
Taylor series for $ e^x $ can be defined as: $ e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+....$
Now, $\lim\limits_{x \to 0}\dfrac{e^x-e^{-x}}{x}=\lim\limits_{x \to 0} \dfrac{[(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+....)-(1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-....)]}{x}$
or, $=\lim\limits_{x \to 0} \dfrac{2x+\dfrac{2x^3}{3!}+\dfrac{2x^5}{5!}+...}{x}$
or, $=\lim\limits_{x \to 0} [2+\dfrac{2x^3}{3!}+\dfrac{2x^4}{4!}+....]$
or, $=2$
