Answer
$2$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{\sin 2x}{x}$
Recall the series for $\sin x=\Sigma_{k =0}^{\infty} (-1)^k \dfrac{(x)^{2k+1}}{(2k+1)!}$
Now, $L=\lim\limits_{x \to 0} \dfrac{\Sigma_{k =0}^{\infty} (-1)^k \dfrac{(x)^{2k+1}}{(2k+1)!}}{x} \\=\lim\limits_{x \to 0} \Sigma_{k =0}^{\infty}
(-1)^k \dfrac{2^{2k+1} x^{2k}}{(2k+1)!} \\=\lim\limits_{x \to 0}
[ (-1)^0 \dfrac{2^{0+1} x^{2(0)}}{(2(0)+1)!} +(-1)^{1} \dfrac{2^{(2)(1)+1} x^{2(1)}}{(2(1)+1)!} +(-1)^2 \dfrac{2^{(2)(2)+1} x^{2(2)}}{(2(2)+1)!} \\=\dfrac{1}{2}\\=\lim\limits_{x \to 0} (2-\dfrac{8x^2}{3!}+\dfrac{2^5x^4}{5!}+......)\\=2-\dfrac{8(0)^2}{3!}+\dfrac{2^5(0)^4}{5!}+.....\\=2$
