Answer
a. Yes. $0$
b. No.
c. No.
Work Step by Step
Use the figure in the Exercise, given the function: $g(x)=\sqrt x sin\frac{1}{x}$
a. Yes. $\lim_{x\to0^+}g(x)=\lim_{x\to0^+}\sqrt x sin\frac{1}{x}=0$ because $\lim_{x\to0^+}\sqrt x =0$ and $sin\frac{1}{x}$ oscillates within $[-1,1]$
b. No. $\lim_{x\to0^-}g(x)$ does not exist because the function is undefined below zero.
c. No. because $\lim_{x\to0^-}g(x)$ does not exist.
