Exercise 4.3.1

Question

Let $g(x)=\sqrt[3]{x}$.
  \begin{enumerate}[label=(\alph*)] 
  \item Prove that $g$ is continuous at $c=0$.
  \item Prove that $g$ is continuous at a point $c 
e 0$. (The identity $a^{3}-b^{3}=$ $(a-b)\left(a^{2}+a b+b^{2}ight)$ will be helpful.)
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Let $\epsilon > 0$ be arbitrary and set $\delta = \epsilon^3$. If $|x-0| < \delta = \epsilon^3$ then taking the cube root of both sides gives $|x|^{1/3} < 1/\epsilon$ and since $(-x)^{1/3} = -(x^{1/3})$ we have $|x|^{1/3} = |x^{1/3}| < \epsilon$. \item We must make $|x^{1/3} - c^{1/3}| < \epsilon$ by making $|x - c|$ small. The identity given allows us to write $$ |x^{1/3} - c^{1/3}| = |x - c| \cdot |x^{2/3} + x^{1/3}c^{1/3} + c^{2/3}| $$ If we choose $\delta < c$ then $0 < |x| < 2|c|$. Keeping in mind that if $a > b > 0$ then $\sqrt[3]{a} > \sqrt[3]{b}$, we can now bound $$ \begin{aligned} |x^{2/3} + x^{1/3}c^{1/3} + c^{2/3}| &\le |x^{2/3}| + |x^{1/3}c^{1/3}| + |c^{2/3}| \ &\le 2^{2/3}|c^{2/3}| + 2^{1/3}|c^{2/3}| + |c^{2/3}| \ &= K \end{aligned} $$ where $K$ is a constant. Then $$ |x^{1/3} - c^{1/3}| \le |x - c| \cdot K $$ Setting $\delta = \frac{\epsilon}{K}$ gives $|x^{1/3} - c^{1/3}| \le \epsilon$ completing the proof. \end{enumerate}

Exercise 4.3.1

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Exercise 4.3.1

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