#### Answer

(a) $\lim_{x\to0^+}f(x)=0$
(b) $\lim_{x\to0^-}f(x)$ does not exist.
(c) $\lim_{x\to0}f(x)$ does not exist

#### Work Step by Step

(a) $\lim_{x\to0^+}f(x)$
Looking at the graph, we cannot see clearly whether as $x\to0$ from the right, $f(x)$ approaches a single value or not.
However, we know that $$-1\le\sin\frac{1}{x}\le1$$ $$-\sqrt x\le\sqrt x\sin\frac{1}{x}\le\sqrt x$$
And $\lim_{x\to0^+}(-\sqrt x)=\lim_{x\to0^+}\sqrt x=\sqrt0=0$
Therefore, according to Sandwich Theorem, $\lim_{x\to0^+}f(x)=0$
(b) $\lim_{x\to0^-}f(x)$ does not exist. $f(x)$ is not defined as $x\lt0$, meaning there are not any values of $f(x)$ as $x$ approaches $0$ from the left.
(c) Since $\lim_{x\to0^+}f(x)$ exists but $\lim_{x\to0^-}f(x)$ does not exist, $\lim_{x\to0}f(x)$ cannot exist.