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Use the definition of continuity to prove that the given function: $h(x) = x^2$ is continuous at every real numbers.

So far for my proof I have:

Let $\epsilon >0$ be given to us. We must show there exists a $\delta >0$ such that $$|x-c| < \delta \Rightarrow |x^2-c^2| < \epsilon.$$

This is where I run into a problem trying to figure out what to choose for $\delta$ to be equal too.

For my scratch work I have:

$|x^2-c^2| = |(x-c)(x+c)| = |x-c||x+c|$ (I know that the part giving me the problem is $|x+c|$)

Let $|x-c| <1 $

Then, $|x+c| = |x-c+2c|\leq |x-c|+2|c| < 1+ |2c|$

This is where I get stuck I do not know where to go from here

Babado
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Yogibear
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    You are almost there. Choose $\delta=\min(1, \frac{\epsilon}{1+|2c|})$. –  Apr 05 '21 at 20:35
  • If $\delta$ is small then $|x-c| < \delta$ means $x\approx c$ and $x+c\approx 2c$ and $|x+c| \approx 2|c|$. Just how approx? Well $x=c$ within a margin of $\pm \delta$ so $|x+c|= 2|c|$ withing the same margin. So $|x+c| < 2|c| + \delta$. – fleablood Apr 05 '21 at 20:53
  • Oh... I see you already figured $|x+ c| < 1 + |2c|$. That's great! That means $|x^2 - c^2| = |x-c||x+c| < \delta(1+2|c|)$ if we desire $ \delta(1+2|c|)\le \epsilon$ that just means $\delta \le \frac \epsilon{1+ 2|c|}$. We just have to make sure we choose a $\delta$ so that both $\delta < 1$ as well. So any $\delta < \min(1, \frac {\epsilon}{1+2|c|}$ will do. – fleablood Apr 05 '21 at 20:58

2 Answers2

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You had found that $$|x^2 -c^2| < |x-c| (1 + 2|c|)$$ for any $c$ and $x$ such that $|x-c| <1$.

Given $\epsilon >0$, define $\delta :=\min\{1,\frac{\epsilon}{1+2|c|}\}$ (Note that denominator in second term is always positive.) We have :

\begin{align} |x-c| < \delta \implies & |x-c| < 1 \, \text{and} \\ & |x-c| < \frac{\epsilon}{1+2|c|} \end{align} This means that \begin{align}|x^2 -c^2| &< |x-c| (1 + 2|c|)\\ &< \frac{\epsilon}{1+2|c|} (1 + 2|c|) = \epsilon \end{align}

Reader Manifold
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If $x- c\approx 0$ then $x \approx c$ and $|x + c| \approx 2|c|$ and as $c$ is constant for evaluating AT $x = c$ we are allowed to (although I prefer not to) to use $c$ if we are attempting to prove $f$ is continuous at $x=c$. (The thing is, and this is the reason I prefer not to, is that we need a different $\delta$ in terms of $c$ for every point we try to evaluate the continuity of $x$ at. But that is acceptable).

SO to put the argument that $x-c \approx 0$ etc. in valid terms of $\delta$....

Claim: If $|x-c| < \delta$ then $|x+c| < 2|c| + \delta$.

Argument:

If $|x-c| < \delta$ then $-\delta < x- c < \delta$ and $c-\delta < x < c+\delta$. SO $2c - \delta < x+c < 2c +\delta$.

If $c \ge 0$ then $-2c - \delta < 2c - \delta < x+c < 2c + \delta$ and $|x+c| < 2|c| +\delta$.

If $c < 0$ then $-2|c| - \delta < x+ c < -2|c| + \delta < 2|c| + \delta$ and $|x+c| < 2|c| + \delta$.

So either way: $|x+c| < 2|c| + \delta$.

Good enough.

.......

So if $|x-c| < \delta$ then $|(x-c)(x+c)| < \delta(2|c| + \delta) = \delta^2 + 2|c|\delta$.

If we assume $\delta \le 1$ then $\delta^2 \le \delta$ and $\delta^2 + 2|c| \delta \le (1+2|c|)\delta$.

So we want for any $\epsilon > 0$ then $(1+2|c|)\delta \le \epsilon$ so let $\delta = \min (1, \frac {\epsilon}{1+ 2|c|})$ and we are golden.

If $|x-c| < \delta$ then $|x^2 -c^2| = |x-c||x+c|< \delta (\delta + 2|c|)\le \delta^2 + 2|c|\delta \le (1+2|c|)\delta \le \epsilon$.

fleablood
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