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I know the most common proof of $\lim_{x\rightarrow c}x^2=c^2$. But I wonder if my alternative proof is valid and correct. Here's my proof.

Let $\varepsilon>0$, want to find a $\delta>0$ such that $\forall x\in\mathbb{R},0<|x-c|<\delta\Rightarrow |x^2-c^2|<\varepsilon$

For the convenience for observation, suppose that $0<|x-c|<\square$, we want to find out which $\square$ is ok, and then know what $\delta$ to pick. Since $|x^2-c^2|<\varepsilon\Leftrightarrow |x+c||x-c|<\varepsilon$ and

\begin{alignat*}{3} &0<|x-c|<\square\\ \Longleftrightarrow &c-\square<x<c+\square&(except\ x=c)\\ \Longleftrightarrow &2c-\square<x+c<2c+\square\qquad&(except\ x=c)\\ \Longrightarrow &|x+c|< |2c|+\square \end{alignat*}

So we see that if we want $|x+c||x-c|<(|2c|+\square)\square<\varepsilon$ to be true, we need to solve a positive solution of the quadratic inequality $\square^2+|2c|\square-\varepsilon<0$, by the relationship between roots and coefficient, we know $\square^2+|2c|\square-\varepsilon=0$ has exactly one positive and one negative root. Hence the solution of the prior inequality is $(-c-\sqrt{c^2+\varepsilon},-c+\sqrt{c^2+\varepsilon})$, where the right endpoint is positive. Thus, we pick $\delta=-c+\sqrt{c^2+\varepsilon}$ to complete the proof.

Eric
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2 Answers2

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The general idea is great, except I'd be careful about the last $\iff$, since the converse isn't necessarily true. It's safer to use a $\Longrightarrow$, since that's all we really need. Indeed, if $c = 7$ and $\square = 1$, then: $$ 13 < x + 1 < 15 \implies -15 < x + 1 < 15 \iff |x + 1| < 15 $$ But the converse is not true, since if $x = 4$, then $|x + 1| < 15$ is true but $13 < x + 1 < 15$ is false.

Here's a cleaned up version of your proof, organized a bit differently:

Given any $\varepsilon > 0$, let $\delta = \sqrt{c^2 + \varepsilon} - |c|$, which is certainly positive. Then if $0 < |x - c| < \delta$, observe that: \begin{align*} |x^2 - c^2| &= |x - c||x + c| \\ &= |x - c||(x - c) + 2c| \\ &\leq |x - c|(|x - c| + |2c|) &\text{by the triangle inequality} \\ &< \delta(\delta + 2|c|) \\ &= (\sqrt{c^2 + \varepsilon} - |c|)(\sqrt{c^2 + \varepsilon} + |c|) \\ &= (c^2 + \varepsilon) - c^2 \\ &= \varepsilon \end{align*} as desired. $~~\blacksquare$

Adriano
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  • Yes, it's due to my carelessness. I had found that and edited before you just posted your answer. – Eric Jul 24 '16 at 10:55
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You have picked $\delta>0$ such that $\delta=-c+\sqrt{c^2+\varepsilon}$

This is a problem. Let c=-5. so $\delta=5+\sqrt{25+\varepsilon}$ substitute $\varepsilon=0.0001$. So $\delta$ is approximately equal to 10. This is where the problem lies. So x can range from -15 to +5. Let x=0. So $x^2$ =0. But 0 does not into (25-$\varepsilon$,25+$\varepsilon$) or (25-0.0001,25+0.0001). So, there is some x satisfying $0<|x-5|<\delta$ for which f(x) does not line in (25-0.0001,25+0.0001). Hence a contradiction. So, we can't use this $\delta$.

Betarup
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  • See it! In which step I made something wrong? – Eric Jul 24 '16 at 11:33
  • $\delta=-c+\sqrt{c^2+\varepsilon}$. The problem is in this step. As the other poster has pointed out, you could have used, $\delta=-|c|+\sqrt{c^2+\varepsilon}$. instead of $\delta=-c+\sqrt{c^2+\varepsilon}$ – Betarup Jul 24 '16 at 11:43
  • Oh, I see! mistake.. Thanks! – Eric Jul 24 '16 at 12:28