Does this prove the following using the delta-epsilon definition?

Question

I wanted to write a formal proof for the following, is there anything wrong with it and is there anything I can improve on?

Answer 1

Better to state it this way.

For any $\epsilon > 0$ there is a $\delta$ so that $|x-a|< \delta \implies |f(x) - L|< \epsilon$.

Let $\gamma = \frac {\delta}5$.

If $|x - \frac a5| < \gamma$ then $|5x - a| < 5\gamma = \delta$. And therefore $|f(5x) - L| < \epsilon$.

Do you see the difference? I couldn't follow yours as I couldn't figure out what $t$ was supposed to be.

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Maybe it will help to do some concrete examples.

$\lim_{x\to 3} x^2 = 9$.

For any $\epsilon > 0$ we can find a $\delta > 0$ so that $|x - 3|<\delta \implies |x^2 - 9| < \epsilon$.

so for example if $\epsilon = .1$ we can find some $\delta$ so that if $3-\delta < x < 3+ \delta$ then we know $|x^2 - 9| < .1$. Well, what is that $\delta$? If $\delta = .1$ then $2.9<x < 3.1$ would imply that $2.9^2 = 8.41 < x^2 < 3.1^2 = 9.61$ ans so, no, $|x-3|<.1 \not \implies |x^2 - 9|< .1$.

But what if we took $\delta = .01$. Then if $2.99 < x < 3.01$ then $8.9 < 8.9401 < x^2 < 9.0601 < 9.1$ so $|x-3|< .01 \implies |x^2 - 9| < .1$ with lots of room to spare.

But that is only one value of $\epsilon$. What if $\epsilon = .01$. Then we'd have to find a different smaller $\delta$.

Well, for any $\epsilon > 0$ no matter how small, if we let $\delta = \min (\frac \epsilon 7, \frac 17)$

Then if $3-\delta < x < 3+\delta \implies 9 -6\delta + \delta^2 < x^2 < 9 + 6\delta +\delta^2$.

If $\epsilon \le 1$ then $\delta = \frac \epsilon 7< \frac 17$ and $\delta^2 < \delta$. So $9- \epsilon=9-7\delta < 9-6\delta < 9- 6\delta + \delta^2 < x^2 < 9+ 6\delta + \delta^2 < 9+ 7\delta = 9 + \epsilon$.

On the other hand if $\epsilon \ge 1$ then $\delta =\frac 17$ and $3-\frac 17 < x < 3+\frac 17 \implies 9-\epsilon \le 8 < 9 - 6\frac 17 + \frac 1{49}<x^2 < 9 +\frac 7 + \frac 1{49} < 10 \le 9 + \epsilon$.

So that does mean whenever $|x-3| < \min (\frac\epsilon 7, \frac 17)$ then $|x^2 - 9| < \epsilon$. SO $\lim\limits_{x\to 3}x^2 = 9$.

That is what limits mean.

So how do we show that $\lim\limits_{x\to \frac 35}(5x)^2 = 9$?

Well, if $|x- \frac 35|<\min(\frac\epsilon {35}, \frac 1{35}$ then

$|5x - 3| < \min(\frac\epsilon 7, \frac 17)$ so

$|(5x)^2 - 9|< \epsilon$.

And that's that.