## Thomas' Calculus 13th Edition

a. Yes. $0$ b. No. c. No.
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.