Proof Verification: Show that $\lim_{x \to c} x^3 =c^3 $ for any $c \in \mathbb{R}$

Question

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

Proof: let $\epsilon > 0$ be given. Choose $\delta =1$ then, $|x-c|<1 \implies |x|-|c| \leq |x-c| < 1 \implies |x|<1+|c|$

Then $|x^3 - c^3|= |(x-c)(x^2 +xc+c^2)| \leq |(x-c)((1+|c|)^2 + (1+|c|)c +c^2)|<\epsilon$

Choose $ \delta =$ $inf{\frac{\epsilon}{|((1+|c|)^2 +(1+|c|)c +c^2)|},1}$

Then, $lim_{x \to c}x^3 =c^3$

Can anyone verify my proof?

Thank you.

Answer 1

Your proof idea is correct, and fairly well written. Only two points, both minor.

• There is a $\delta$ in the starting which you set to $1$ and then you reset it later to $\inf\{...,1\}$. You can avoid multiple assignment to the same variable by either using different variables, such as $\delta_0$ etcetera.

• The $\delta$ that you obtain in the end must not be infinite, which is not entirely obvious from its form. Thankfully, you can check that the denominator has some sum of square components, which cannot be zero.