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## Sep 14 Proving limits of square roots

Let's say we have this problem:

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

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

MathJax TeX Test Page $$|2\sqrt{x} + 3 - 7| < \epsilon$$ $$|2\sqrt{x} - 4| < \epsilon$$ $$2|\sqrt{x} - 2| < \epsilon$$ 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 bound for delta. Let's say it's 2. It can be anything though. 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

MathJax TeX Test Page Since we set the bound for delta from 2 to 6, we can write:
Let $$\delta = \min\left(\dfrac{\sqrt{2} + 2}{2} * \epsilon, 2\right)$$ $$|x-4| < \delta$$ Now, $\delta \leq \dfrac{\sqrt{2} + 2}{2}$, so we can rewrite the last line as: $$|x-4| < \dfrac{\sqrt{2} + 2}{2} * \epsilon$$ Now, we multiply and extend the inequality $$2\dfrac{|x-4|}{\sqrt{x} + 2} \leq 2\dfrac{|x-4|}{\sqrt{2} + 2} < \epsilon$$ By the transitive property, $$2\dfrac{|x-4|}{|\sqrt{x} + 2|} < \epsilon$$ $$2|\sqrt{x} - 2| < \epsilon$$ $$|2\sqrt{x} - 4| < \epsilon$$ $$|2\sqrt{x} + 3 - 7| < \epsilon$$ $$|f(x) - L| < \epsilon$$
David Witten