Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition

Question

$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$

Let $\epsilon>0$. Then $$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$

Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$. Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$, $$|x^3-c^3|=|x-c||x^2+xc+c^2|<ε.$$

Does this make sense or are the steps done in the right way?

Answer 1

Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-c|<\delta$, then $|x^3-c^3|<\epsilon$. Now, $$ |x^3-c^3| = |x-c||x^2+xc+c^2|. $$ If $|x-c|<1$, then we have that $-1<x-c<1$ or simply $c-1<x<c+1$ so that $$ x^2+xc+c^2<(c+1)^2+(c+1)(c)+c^2=(c^2+2c+1)+(c^2+c)+c^2=3c^2+3c+1, $$ and so $$ |x^3-c^3| = |x-c||x^2+xc+c^2|<(3c^2+3c+1)|x-c|. $$ So if we take $\delta=\min\left\{1,\frac{\epsilon}{3c^2+3c+1}\right\}$, then $0<|x-c|<\delta$ implies that $$ |x^3-c^3|=|x-c||x^2+xc+c^2|<\frac{\epsilon}{3c^2+3c+1}\cdot(3c^2+3c+1)=\epsilon. $$ Thus, by the definition of a limit, we have that $$ \lim_{x\to c}x^3=c^3. \blacksquare $$

Answer 2

Hint: $\delta$ should not be depending of $x$.

Consider $c > 0$ then

$$ c - 1 < x < c+ 1 \implies |x| < c + 1 $$

Then $$\begin{align}|x^3 - c^3| &= |x - c| |x^2 + xc +c^2| \\&\leq |x-c|\bigg(|c+1|^2 + |c+1||c| + + |c|^2\bigg) \\&=|x-c|\bigg[|c+1|(|c + 1| + |c|) + |c|^2\bigg]\end{align} $$

Take $$\delta = \min \Bigg\{1, \frac{\epsilon}{|c+1|(|c + 1| + |c|) + |c|^2}\Bigg\}$$

Answer 3

Suppose $\varepsilon>0$. The inequality $$ |x^3-c^3|<\varepsilon $$ becomes $$ c^3-\varepsilon<x^3<c^3+\varepsilon $$ or $$ \sqrt[3]{c^3-\varepsilon}<x<\sqrt[3]{c^3+\varepsilon} $$ or also $$ \sqrt[3]{c^3-\varepsilon}-c<x-c<\sqrt[3]{c^3+\varepsilon}-c $$ Take $$ \delta=\min\{c-\sqrt[3]{c^3-\varepsilon},\sqrt[3]{c^3+\varepsilon}-c\} $$ after noting that $$ c-\sqrt[3]{c^3-\varepsilon}>0 $$ and $$ \sqrt[3]{c^3+\varepsilon}-c>0 $$

Answer 4

Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-c|<\delta$, then $|x^3-c^3|<\epsilon$.

Now, $ |x^3-c^3| = |x-c||x^2+cx+c^2| $

If $|x-c|<1$, then $||x|-|c|| \leq |x-c|<1$

i.e. $||x|-|c||<1 \implies -1<|x|-|c|<1 \implies |x|<|c|+1$ so that $$ |x^2+cx+c^2|\leq |x|^2+|c||x|+|c|^2 < (|c|+1)^2+|c|(|c|+1)+|c|^2=3|c|^2+3|c|+1, $$ and so $$ |x^3-c^3| = |x-c||x^2+cx+c^2|<(3|c|^2+3|c|+1)|x-c|. $$ Therefore, for every $\epsilon >0$, there exists a $\delta=\inf\left\{1,\frac{\epsilon}{3|c|^2+3|c|+1}\right\}>0$, such that if $ 0<|x-c|<\delta$ then

$\begin{align} |x^3-c^3|&=|x-c||x^2+cx+c^2|\\ &<\frac{\epsilon}{3|c|^2+3|c|+1}\cdot(3|c|^2+3|c|+1)\\ &=\epsilon \end{align} $

Thus, by the definition of limit of a function $$ \lim_{x\to c}x^3=c^3. $$