Answer
$\dfrac{7}{2}$
Work Step by Step
We know that the Taylor Series for $e^x$ and $\sin x$ is defined as:
$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ and $ \sin x={(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Now, $\lim\limits_{ x \to 0}\dfrac{7 \sin x}{e^{2x}-1}=\lim\limits_{ x \to 0}\dfrac{7 (x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)}{(1+2x+\dfrac{(2x)^2}{2!}+\dfrac{(2x)^3}{3!}+\dfrac{(2x)^4}{4!}+.)-1}$
or, $=\lim\limits_{ x \to 0}[\dfrac{7x (1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-...)}{x(2+\dfrac{4x}{2!}+\dfrac{8x^2}{3!}+...)}]$
or, $\dfrac{7 (1-0+0-...)}{(2+0+0+...)}=\dfrac{7}{2}$
