# $\epsilon$ - $\delta$ proof for continuity of $\lim_{x \to a} \sqrt{x}$ for $a > 0$

I am currently undergoing a Calculus I course as an undergraduate. I was hoping to gain some guidance into whether the proof I have worked out for the continuity of a function $$f(x) = \sqrt{x}$$ is robust enough or is missing out on certain details.

Aim: Prove that the function $$f(x) = \sqrt{x}$$ is continuous at $$a > 0$$.

(1) Finding the value of f(a) for which f is defined

The function $$f(x) = \sqrt{x}$$ is defined for values of $$x \ge 0$$. Given $$a > 0$$,

$$f(a) = \sqrt{a} \\ \lim_{x \to a} \sqrt{x} = \sqrt{a}$$

(2) Showing that $$\lim\limits_{x \to a} \sqrt{x}$$ exists for every $$a > 0$$.

Hence, given $$\epsilon > 0$$, we aim to find a proper $$\delta > 0$$ such that:

$$0 < |x - a| < \delta \Rightarrow |\sqrt{x} - \sqrt{a}| < \epsilon$$

Factorizing $$|x-a|$$,

$$|x-a| = |\sqrt{x} - \sqrt{a}||\sqrt{x} + \sqrt{a}| < \delta \\ |\sqrt{x} - \sqrt{a}| < \frac{\delta}{|\sqrt{x} + \sqrt{a}|}$$

Since $$\sqrt{x} + \sqrt{a}\ge 0$$, it follows that $$\frac{1}{|\sqrt{x} + \sqrt{a}|} < 1$$. Hence,

$$|\sqrt{x} - \sqrt{a}| < \frac{\delta}{|\sqrt{x} + \sqrt{a}|} < \delta$$

Therefore, we take $$\delta = \epsilon$$.

Proof:

$$0 < |x - a| < \delta \Rightarrow |\sqrt{x} - \sqrt{a}| < \delta = \epsilon.$$

Since $$f(a) = \sqrt{a}$$ and $$\lim\limits_{x \to a} \sqrt{x} = \sqrt{a}$$ for $$a > 0$$,

$$\lim_{x \to a} \sqrt{x} = f(a)$$

and thus the function $$f(x) = \sqrt{x}$$ is continuous at every a > 0.

Any feedback on gaps or loopholes in this proof would be greatly appreciated!

Edit:

There was a flaw in my logic, as kindly pointed out below. Since $$\sqrt{x} + \sqrt{a}$$ may fall in the range of values $$0 \le \sqrt{x} + \sqrt{a} \le 1$$ the assertion that $$\frac{1}{|\sqrt{x} + \sqrt{a}|}$$ does not hold.

The appropriate logic should be:

$$\frac{\delta}{|\sqrt{x} + \sqrt{a}|} < \frac{\delta}{|\sqrt{a}|}$$

for x > 0.

Hence,

$$|\sqrt{x} - \sqrt{a}| < \frac{\delta}{|\sqrt{a}|}$$

We choose $$\delta = \epsilon\sqrt{a}$$.

Revised proof:

$$0 < |x - a| < \delta \Rightarrow |\sqrt{x} - \sqrt{a}| < \epsilon\sqrt{a}(\sqrt{a}) = \epsilon$$

• Close. But $\sqrt{x} + \sqrt{a} \geq 0$ does not imply $1/|\sqrt{x} + \sqrt{a}| < 1$. – aschepler Aug 22 '20 at 12:13
• RIGHT! Thank you! Must have messed up it up by assuming that $a \ge 1$. – iobtl Aug 22 '20 at 13:14

There is no reason why $$\frac {\delta} {\sqrt x+\sqrt a}$$ should be less than $$\delta$$ so your proof is not valid.

Note that $$\frac {\delta} {\sqrt x+\sqrt a} < \frac {\delta} {\sqrt a}$$ (for $$x>0$$) so it is enough to choose $$\delta =\epsilon \sqrt a$$

• I realize my mistake. Thank you for the correction! – iobtl Aug 22 '20 at 13:15

Let $$\varepsilon>0$$ be defined. We want to find such $$\delta>0$$ that suffices $$\forall x\in (a-\delta,a+\delta), |f(x)-f(a)|<\varepsilon$$. So, we're looking for such $$\delta$$ that: $$|\sqrt x-\sqrt a|<\varepsilon$$. Notice: $$|\sqrt x-\sqrt a|=\frac{|x-a|}{\sqrt x + \sqrt a}<\frac{|x-a|}{\sqrt a}<\frac{\delta}{\sqrt a}$$.

Now let's define $$\delta = \varepsilon \sqrt a$$.

Therefore: $$|\sqrt x-\sqrt a|=\frac{|x-a|}{\sqrt x + \sqrt a}<\frac{|x-a|}{\sqrt a}<\frac{\delta}{\sqrt a}=\frac{\varepsilon \sqrt a}{\sqrt a}=\varepsilon$$.

** Notice that $$\sqrt x, \sqrt a>0$$ so the expression $$\sqrt x + \sqrt a$$ doesn't need absolute value.