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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.

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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.

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  • $\begingroup$ Thank you! I understand the second point. However, what do you mean by the first bullet point? $\endgroup$
    – Alea
    Apr 11 '18 at 6:01
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    $\begingroup$ He's saying that you already chose the variable $\delta$ to equal 1, but then you use the same variable to equal the infinum. To avoid this, let $\delta_1=1$. Then let $\delta_2=\inf...$. That's all. $\endgroup$ Apr 11 '18 at 12:05
  • $\begingroup$ Oh! Yes, I get it now. If I've already said that $\delta =1$ then $\delta = inf{..,1}$ makes no sense at all. Thanks. $\endgroup$
    – Alea
    Apr 11 '18 at 19:13

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