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We want to prove $f(x) =x^2$ is continuous at $x_0=2$ using the $\epsilon$-$\delta$ definition.

  • My attempt:

We want the function $f$ to satisfy the definition of continuity, meaning :

For all $\epsilon \gt 0$ there exists $\delta$ such that at $x_0=2$, $ \lvert x-2 \rvert \lt \delta \implies \lvert f(x) - 4 \rvert \lt \epsilon$

Notice that $f(x)$ is really just $x^2$. Substituting this in we get $ \lvert x^2 - 4 \rvert \lt \epsilon$.

$ \lvert x^2 - 4 \rvert = \lvert x-2 \rvert \lvert x+2 \rvert \lt \epsilon$.

This is where I get stuck:

  • How do we recognize the correct choice of $\epsilon$ and $\delta$ at this point?

Thank you!

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  • $\begingroup$ This is a good start. A suggestion to make your writing smoother: you don't need to write "Notice that $f(x)$ is really just $x^2$." There's nothing noteworthy here, since this is the definition of $f$. $\endgroup$ – Théophile May 14 '15 at 22:53
  • $\begingroup$ @Théophile, I'm starting to realize this as I post on here more and more. People don't seem to like wordy proofs that state redundant things. Who would've known ;P. Thanks $\endgroup$ – David South May 14 '15 at 22:54
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Note that $|x+2|<5$ for $|x-2| < 1$. So lets take $|x-2| < 1$. Then you have

$$|x-2||x+2| < |x-2|5$$

If $|x-2| < \tfrac \epsilon5$ you get $|x-2|5 < \epsilon$. We want

  • $|x-2| < 1$
  • $|x-2| < \tfrac\epsilon5$

Thus we take $\delta = \min\{1,\tfrac\epsilon5\}$, because if $|x-2|<\min\{1,\tfrac\epsilon5\}$ then both of the above conditions are fulfilled.

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  • $\begingroup$ It should be $\delta\le1$ or at the end you should use, say, $1/2$ and not $1$. Just being picky, of course. $\endgroup$ – egreg May 14 '15 at 22:40
  • $\begingroup$ @egreg, why is it better to use ½ and not 1? $\endgroup$ – David South May 14 '15 at 22:41
  • $\begingroup$ @DavidSouth In the first line the assumption is made that $\delta<1$, whereas $\min\{1,\epsilon/5\}$ might be $1$. But it's easily fixable. $\endgroup$ – egreg May 14 '15 at 22:42
  • $\begingroup$ @egreg, ok thank you! that makes sense $\endgroup$ – David South May 14 '15 at 22:43
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    $\begingroup$ I don't think it's necessary to multiply by $\frac12$; it's enough to take $\delta = \min\{1, \frac\epsilon 5\}$, because we're assuming that $|x-2| < \delta$. $\endgroup$ – Théophile May 15 '15 at 3:06

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