Answer
$1$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to \infty} x \sin x \dfrac{1}{x}$
Recall the series for $\sin x=\Sigma_{k =0}^{\infty} (-1)^k \dfrac{(x)^{2k+1}}{(2k+1)!}$
Now, $L=\lim\limits_{t \to 0}\dfrac{1}{t} \sin t\\=\lim\limits_{t \to 0}\dfrac{1}{t} \Sigma_{k =0}^{\infty} (-1)^k \dfrac{(t)^{2k+1}}{(2k+1)!}\\=\lim\limits_{t \to 0}[(-1)^0 \dfrac{(t)^{(2)(0)}}{(2(0)+1)!}+(-1)^1 \dfrac{(t)^{(2)(1)}}{(2(1)+1)!}+(-1)^2 \dfrac{(t)^{(2)(2)}}{(2(2)+1)!}+.........)\\=1-\dfrac{0^2}{3!}+\dfrac{0^4}{5!}+..........\\=1$
