Evaluating an infinity times zero limit via L'Hopital's Rule

The original limit has the form $\infty \cdot 0$, which is not directly a form L'Hopital's Rule applies to. Rewrite it as a quotient: $x\sin(1/x) = \dfrac{\sin(1/x)}{1/x}$. As $x \to \infty$, $\sin(1/x) \to 0$ and $1/x \to 0$, so this is now the indeterminate form $0/0$, and L'Hopital's Rule applies.

Differentiate numerator and denominator separately: 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 $-1/x^2$. So the limit becomes $\lim_{x\to\infty} \dfrac{-\cos(1/x)/x^2}{-1/x^2} = \lim_{x\to\infty} \cos(1/x)$. As $x \to \infty$, $1/x \to 0$, so $\cos(1/x) \to \cos(0) = 1$.

The original limit was infinity times zero. As x approaches infinity, x grows without bound while sine of one over x approaches zero, giving this indeterminate product form. This form cannot be handled directly by L'Hopital's Rule, which is why the expression needed to be rewritten as a quotient first. Rewriting it as sine of one over x divided by one over x converted it into the zero over zero form.