Proving limits of square roots

See also: Proving Limits of Quadratics

Let's say we have this problem:

$$\lim_{x \to 4} \left(2\sqrt{x} + 3\right) = 7$$

Recall that we want to find a relationship between delta (distance x is from 4) and epsilon (distance the function is from 7). Here, we're going to be doing an epsilon-delta proof of this square root function.

Discovery Phase

$$|2\sqrt{x} + 3 - 7| < \epsilon$$ $$|2\sqrt{x} - 4| < \epsilon$$ $$2|\sqrt{x} - 2| < \epsilon$$ Now, $|x - 4| = |\sqrt{x} + 2||\sqrt{x} - 2|$, so we can rewrite it. $$2\frac{|x-4|}{|\sqrt{x} + 2|} < \epsilon$$ Now, we have to set a maximum value for delta. Let's say it's 2. For really big values of $\epsilon$, $\delta = 2$ still works. For small values of $\epsilon$, $\delta$ is unaffected. So, x must be in $[c - 2, c + 2] = [2,6]$ We can write this inequality: $$2\frac{|x-4|}{|\sqrt{2} + 2|} \geq 2\frac{|x-4|}{|\sqrt{x} + 2|} \geq 2\frac{|x-4|}{|\sqrt{6} + 2|}$$ So, if we prove $\epsilon > 2\dfrac{|x-4|}{\sqrt{2} + 2}$, we prove the general case.

Proof

We are trying to prove this statement: $$\forall \epsilon > 0, \exists \delta \text{ }s.t. \text{ if } 0 < |x-4| < \delta \to |2\sqrt{x} - 4| < \epsilon$$ In order to do this, we show that it works for an arbitrary $\epsilon > 0$. $$\text{Let }\epsilon > 0$$ This leaves us with two cases: $\dfrac{\sqrt{2} + 2}{2} * \epsilon > 2$ or $\leq 2$ $$\text{Case 1: } \epsilon > \dfrac{4}{\sqrt{2} + 2}$$ $$\text{Let }\delta = \min\left(\dfrac{\sqrt{2} + 2}{2} * \epsilon, 2\right) = 2$$ $$|x-4| < 2 \to \dfrac{2|x-2|}{\sqrt{2} + 2} \lt \dfrac{4}{\sqrt{2} + 2}$$ Because $x \in [4 - \delta, 4 + \delta] = [2,6]$, $$\dfrac{2|x-4|}{|\sqrt{x} + 2|} \leq \dfrac{2|x-2|}{\sqrt{2} + 2} \lt \dfrac{4}{\sqrt{2} + 2}$$ $$\dfrac{2|x-4|}{|\sqrt{x} + 2|} = 2|\sqrt{x} - 2| = \boxed{|2\sqrt{x} + 3 - 7| \lt \dfrac{4}{\sqrt{2} + 2} < \epsilon}$$ Now, that case is done. Now, we do the unaffected case, where $\epsilon \leq \dfrac{4}{2 + \sqrt{2}}$ $$\text{Let } \delta = \min\left(\dfrac{\sqrt{2} + 2}{2} * \epsilon, 2\right) = \dfrac{\sqrt{2} + 2}{2} \epsilon$$ $$|x - 4| < \dfrac{\sqrt{2} + 2}{2} \epsilon$$ $$\dfrac{2|x-4|}{\sqrt{2} + 2} < \epsilon$$ Because $x \in [2, 6]$, we can write $$\dfrac{2|x-4|}{\sqrt{x} + 2} < \dfrac{2|x-4|}{\sqrt{2} + 2} < \epsilon$$ $$\boxed{|\sqrt{x} + 3 - 7| < \epsilon}$$ For every epsilon, we have now proven that there exists a delta such that if $|x-4|< \delta$, then $|\sqrt{x} + 3 - 7| < \epsilon$.