Correct answer for this limit?

Question

So I tutor this student, and we came to the conclusion that the delta needs to be 0.1975, but her TA thinks it should be 0.2025. This was a homework question for her, and she got it wrong because she originally had what I had. Where did I go wrong, and how can you prove it using the epsilon delta proof? Everything I have found, and every attempt at a solution I give says that we were correct...

I found this answer by finding b and c, then finding the distances from x=4. Knowing the nature of $\sqrt{x}$, the shortest distance would need to be delta to avoid being outside the range of epsilon by picking the larger distance. This comes out to the distance from b to 4, which is 0.1975.

I proved this using the standard epsilon delta proof and it worked out. If needed, I can go through my steps here

Answer 1

Sanity check:

$$\begin{align}|\sqrt{4-0.1975}-2|&=\color{green}{0.05} \\|\sqrt{4+0.1975}-2|&=\color{green}{0.04878\cdots} \\|\sqrt{4-0.2025}-2|&=\color{red}{0.06415\cdots} \\|\sqrt{4+0.2025}-2|&=\color{green}{0.05}\end{align}$$

Answer 2

The TA's answer is clearly incorrect, because if $\delta = 0.2025$, you can choose $x = 3.8$, which satisfies the condition $|x - 4| < \delta$, but not $|\sqrt{x} - 2| < 0.05$. If the TA cannot accept this simple counterexample, then they are not competent to grade this question.

Answer 3

Your answer is correct!

Note: $$|x-4|<\delta \Rightarrow 4-\delta <x<4+\delta \Rightarrow \sqrt{4-\delta}<\sqrt{x}<\sqrt{4+\delta}\\ |2-\sqrt{x}|=\big{|}\frac{(2-\sqrt{x})(2+\sqrt{x})}{2+\sqrt{x}}\big{|}=|\frac{4-x}{2+\sqrt{x}}|<\frac{\delta}{2+\sqrt{4-\delta}}=\epsilon=0.05 \Rightarrow \delta=0.1975.$$