Correct way of finding $\delta $ for $\lim_{x \to a} \sqrt{x} = \sqrt{a}$

Question

Prove: $\lim_{x \to a} \sqrt{x} = \sqrt{a}$ using $\epsilon$-$\delta$.

My solution:

We have that $0 < |x-a| < \delta $.

Also, $ |\sqrt{x} - \sqrt{a}| < \epsilon $

$ \therefore -\epsilon < \sqrt{x} - \sqrt{a} < \epsilon $

$ \therefore \sqrt{a} -\epsilon < \sqrt{x} < \sqrt{a} + \epsilon $

$ \therefore (\sqrt{a} -\epsilon)^2 < x < (\sqrt{a} + \epsilon)^2 $

$ \therefore (\sqrt{a} -\epsilon)^2 - a < x - a < (\sqrt{a} + \epsilon)^2 - a $

$ \therefore \epsilon^2 - 2\sqrt{a}\epsilon < x - a < \epsilon^2 + 2\sqrt{a}\epsilon $

$ \delta = \min(|\epsilon^2 - 2\sqrt{a}\epsilon|,|\ \epsilon^2 + 2\sqrt{a}\epsilon|) $

By letting $ \delta = \min(|\epsilon^2 - 2\sqrt{a}\epsilon|,|\ \epsilon^2 + 2\sqrt{a}\epsilon|) $, we get that $|\sqrt{x}-\sqrt{a}|<\epsilon$ if $0 < |x-a| < \delta $.

Thus, $\lim_{x \to a} \sqrt{x} = \sqrt{a}$.

Question: is it correct to find a $ \delta $ using the method above?

I've seen a bunch of other methods used to find $\delta$ but I'm totally confused what is the "right" way of doing it. If anyone could explain this or give some other examples, I would appreciate it.

Thank you a lot!

Answer 1

That certainly works. There isn't really a "right" way to do these, but some approaches are shorter than others. In some instances it may be important to find the largest possible $\delta$, but to verify the definition of limit you only need to demonstrate one. For instance, if $a > 0$ you can use $x-a = (\sqrt x - \sqrt a)(\sqrt x + \sqrt a)$ to get $$|\sqrt x - \sqrt a| = \frac{|x-a|}{|\sqrt x + \sqrt a|} \le \frac{|x-a|}{\sqrt a}.$$ Thus if $\delta = \epsilon \sqrt a$ then $|x-a| < \delta$ implies $|\sqrt x - \sqrt a| < \epsilon$.

Answer 2

There's a bit of an issue: if $\epsilon=2\sqrt{a},$ then you would have $\delta=0.$ Now, in the case that $a=0,$ this is impossible, and you'll instead have $\delta=\epsilon^2.$ In the case that $a>0,$ you should start by assuming that $0<\epsilon<2\sqrt{a},$ which will fix the problem. You will have to justify that it's okay to only consider such $\epsilon,$ however.

Answer 3

The OP's logic can be salvaged. One problem is they assume on their third $\therefore$ that $u \lt v$ implies $u^2 \lt v^2$ without knowing for sure that $u \ge 0$.

Here is a 'logic patch':

If $a = 0$, simple algebra shows that $\delta = \varepsilon^2$ works.

To show continuity when $a \gt 0$, we can take a few easy steps:

$\tag 1 |\sqrt{x} - \sqrt{a}| < \varepsilon $

$ \therefore -\varepsilon < \sqrt{x} - \sqrt{a} < \varepsilon $

$ \therefore \sqrt{a} -\varepsilon < \sqrt{x} < \sqrt{a} + \varepsilon $

Now to get rid of $\sqrt x$ and replace it with $x$, we would like to run the 'square it' on our expression, but to do that we must assume that $\sqrt{a} -\varepsilon$ in non-negative. i.e. $\varepsilon \le \sqrt{a}$. But that is not a restriction at all, so we proceed:

$ \therefore (\sqrt{a} -\varepsilon )^2 < x < (\sqrt{a} + \varepsilon )^2 $

$ \therefore -2 \sqrt{a} \, \varepsilon + \varepsilon^2 < x - a < 2 \sqrt{a} \, \varepsilon + \varepsilon^2$

We are trying to find our $\delta \gt 0$ 'setup':

$\tag 2 -2 \sqrt{a} \, \varepsilon + \varepsilon^2 \le -\delta < x - a < \delta \le 2 \sqrt{a} \, \varepsilon + \varepsilon^2$

so that $\text{(2)}$ implies $\text{(1)}$ (we can use our developed "$\therefore$-logic-chain" in reverse).

Since $\varepsilon \le \sqrt{a}$ here, we can define our $\delta \gt 0$ with

$\tag 3 \delta = 2 \sqrt{a} \, \varepsilon - \varepsilon^2$

and so by definition the left side of $\text{(2)}$ 'clicks'. For the right side, it is easy to see that $\delta \le 2 \sqrt{a} \, \varepsilon + \varepsilon^2$ since

$\quad (2 \sqrt{a} \, \varepsilon - \varepsilon^2) \le (2 \sqrt{a} \, \varepsilon + \varepsilon^2) \text{ iff } 2 \varepsilon^2 \ge 0$

So we can use the OP's logic to find a $\delta$ given any starting $\varepsilon$, where if necessary we pound $\varepsilon$ down first to $\sqrt a$.