L'Hopital's Rule with a rewritten indeterminate form

Rewrite the product as a quotient: $x\sin(1/x) = \dfrac{\sin(1/x)}{1/x}$. As $x \to \infty$, $1/x \to 0$, so $\sin(1/x) \to 0$ and $1/x \to 0$, giving the indeterminate form $0/0$. This confirms L'Hopital's Rule can be applied.

Differentiate numerator and denominator separately with respect to $x$: the derivative of $\sin(1/x)$ is $\cos(1/x)\cdot(-1/x^2)$ by the chain rule, and the derivative of $1/x$ is $-1/x^2$. Forming the new quotient, $\lim_{x\to\infty} \frac{-\cos(1/x)/x^2}{-1/x^2} = \lim_{x\to\infty} \cos(1/x).$ As $x \to \infty$, $1/x \to 0$, so $\cos(1/x) \to \cos(0) = 1$. Therefore the original limit equals $1$.

The answer is one. After differentiating numerator and denominator, the quotient simplified to cosine of one over x. As x approaches infinity, one over x approaches zero, so cosine of one over x approaches cosine of zero, which equals one.