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 states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
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, $\lim\limits_{x \to 0} \dfrac{x^2-2x}{x^2-\sin x}=\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x} \ne \lim\limits_{x \to 0} \dfrac{2}{2+\sin x}=1$
(L'Hospital's rule can not be used because we do not have indeterminate form.)
b) $\lim\limits_{x \to 0} \dfrac{x^2-2x}{x^2-\sin x}=\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x}=\dfrac{2(0)-2}{2(0)-\cos 0}=2$
Hence,
a) The limit provided in part(a) is not correct.
b) The limit provided in part(b) is correct.