Answer
a) The limit provided in part(a) is not correct.
b) The limit provided in part(b) is correct.
Work Step by Step
(a) L-Hospital's rule is defined as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
Here, $\lim\limits_{x \to 0} \dfrac{x^2-2x}{x^2-\sin x}=\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x}=1$
Also, $\lim\limits_{x \to 0} \dfrac{x^2-2x}{x^2-\sin x}=\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x} $
and $\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x} \ne \lim\limits_{x \to 0} \dfrac{2}{2+\sin x}=1$
We cannot use L-Hospital's rule because we do not get any indeterminate form of limit.
(b) $\lim\limits_{x \to 0} \dfrac{x^2-2x}{x^2-\sin x}=\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x}$
or, $\dfrac{0-2}{0-1}=2$
