Limit of $\sqrt{x}$ as $x$ approaches $a$. Delta-epsilon.

Question

Find $$\lim_{x \to a} \sqrt x $$ using $\varepsilon-\delta$ definition.

Could someone suggest a hint ?

I'm having trouble advancing beyond the starting line

$$|x -a| < \delta \quad \Rightarrow \quad |\sqrt x \ - \sqrt a | < \varepsilon$$

Because I don't know how to put $|\sqrt x \ - \sqrt a |$ into the form of $|x -a|$

P.S. How do you vertically space with MathJAx? I tried using \vspace (Latex) to no avail...

Answer 1

I'm going to give you the general outline for any proof of this type. I'll quote the bits that you'd write down, and put my explanation/comments in normal text. I'm also going to assume that $a > 0$, to make everything stay real and well-defined.

For any $\varepsilon > 0$, let $\delta = $

We don't know what $\delta$ needs to be yet, so we'll just leave a space there. For a start, though, we're going to need $x > 0$, so we'll stick $\delta \leq a$ in our list of things that we need.

Then for any $x$ such that $|x-a|<\delta$, we have $$|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}}$$

That $\sqrt{x}$ on the bottom is annoying, so we want to get rid of it. In particular, we want an upper bound on the whole fraction, so we want a lower bound on the denominator, hence a lower bound on $\sqrt{x}$. It doesn't matter what we pick, so long as it bounds $\sqrt{x}+\sqrt{a}$ away from $0$, but that's easy: $\sqrt{x} > 0$ already does that, so we can just use it again:

$$\leq \frac{|x-a|}{\sqrt{a}} $$

And, by definition of $x$, we have

$$< \frac{\delta}{\sqrt{a}}$$ And now we know what else we need out of $\delta$: we need $\frac{\delta}{\sqrt{a}}\leq\varepsilon$, so $\delta\leq \varepsilon\sqrt{a}$. Once we've assumed that, we've got

$$\leq \varepsilon$$

exactly as we wanted. So, we need to go and fill in the gap at the top with what needs to be there: we know that we need $\delta \leq a$ and $\delta \leq \varepsilon\sqrt{a}$, so we'll take the obvious answer and set $\delta = \min\{a,\varepsilon\sqrt{a}\}$, and go and fill that in.

So, putting all of that together and sticking the standard boilerplate on the end, we've got:

For any $\varepsilon > 0$, let $\delta = \min\{a,\varepsilon\sqrt{a}\}$. Then for any $x$ such that $|x-a| < \delta$, we have \begin{align*}|\sqrt{x}-\sqrt{a}|&=\frac{|x-a|}{\sqrt{x}+\sqrt{a}}\\&\leq \frac{|x-a|}{\sqrt{a}}&\mbox{since }\sqrt{x}\geq 0\\&<\frac{\delta}{\sqrt{a}}&\mbox{since }|x-a|<\delta\\&\leq\varepsilon&\mbox{since }\delta \leq \varepsilon\sqrt{a}\end{align*} so $$\lim\limits_{x\rightarrow a}\sqrt{x}=\sqrt{a}.$$