## Thomas' Calculus 13th Edition

(a) $0$ (b) $0$ (c) $0$
Given the piecewise function $f(x)=\begin{cases} x^2sin(1/x)\hspace1cm x\lt0 \\ \sqrt x\hspace2.5cm x\gt0 \end{cases}$ (a) For $x\to0^+$, we have $\lim_{x\to0^+}\sqrt x=0$. To verify this, for any $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ in the interval of $0\lt x\lt \delta$, we get $|\sqrt x-0|\lt\epsilon$. The last inequality gives $0\lt \sqrt x\lt\epsilon$ or $0\lt x\lt\epsilon^2$, thus if we choose $\delta=\epsilon^2$ and go backwards with the procedure, we have for all $x$ in the interval of $0\lt x\lt \delta$, we get $|\sqrt x-0|\lt\epsilon$ which proves the limit statement. (b) As function $sin(1/x)$ oscillates in a range of $[-1,1]$, we have $-x^2\leq x^2sin(1/x)\leq x^2$, use the Sandwich Theorem, we have $\lim_{x\to0^-}x^2sin(1/x)=\lim_{x\to0^-}x^2=0$. To verify this, use a similar procedure in part (a). For any $\epsilon\gt0$, we need to find a value $\delta\gt0$, for all $x$ values in the interval of $-\delta\lt x\lt 0$, we have $|x^2-0|\lt\epsilon$ which requires $|x|\lt\sqrt {\epsilon}$ or $-\sqrt {\epsilon}\lt x\lt 0$. Thus we can choose $\delta=\sqrt {\epsilon}$ so that for all $x$ in the interval of $-\delta\lt x\lt 0$, we get $|x^2-0|\lt\epsilon$ which proves the limit statement. (c) Because $\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=0$, the limit $\lim_{x\to0}f(x)=0$