Answer
$\dfrac{7}{2}$
Work Step by Step
Find $\lim\limits_{ x \to 0}\dfrac{7 \sin x}{e^{2x}-1}$
Now, $\lim\limits_{ x \to 0}\dfrac{7 \sin x}{e^{2x}-1}=\lim\limits_{ x \to 0}\dfrac{7 (x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)}{(1+2x+\dfrac{2^2x^2}{2!}+\dfrac{2^3x^3}{3!}+\dfrac{2^4x^4}{4!}+.)-1}$
and $\lim\limits_{ x \to 0}[\dfrac{7x (1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-...)}{x(2+\dfrac{4x}{2!}+\dfrac{8x^2}{3!}+...)}]=\dfrac{7 (1-0+0-...)}{(2+0+0+...)}=\dfrac{7(1)}{2}=\dfrac{7}{2}$
