Answer
$a)$ $\infty\times0$
$b)$ $1$
$c)$ see picture
Work Step by Step
$a)$$\lim\limits_{x \to \infty}xsin\frac{1}{x}=(\infty)sin(\frac{1}{\infty})=(\infty)sin0=\infty\times0$
$\infty\times0$ is an indeterminate form.
$b)$ $\lim\limits_{x \to \infty}xsin\frac{1}{x}=(\infty)sin(\frac{1}{\infty})=(\infty)sin0=\infty\times0$
$\infty\times0$ is an indeterminate form. We can't do l'Hopital's Rule until the indeterminate form is either $\frac{0}{0}$ or $\frac{\infty}{\infty}$, so:
$\lim\limits_{x \to \infty}\frac{sin\frac{1}{x}}{\frac{1}{x}}=\frac{sin\frac{1}{\infty}}{\frac{1}{\infty}}=\frac{sin0}{0}=\frac{0}{0}$
Now we can do l'Hopital's Rule:
$\lim\limits_{x \to \infty}(\frac{cos\frac{1}{x}(-x^{-2})}{-x^{-2}})$
$\lim\limits_{x \to \infty}cos\frac{1}{x}=cos\frac{1}{\infty}=cos0=1$
$c)$ see picture
