Given $\epsilon$, find $\delta$

Question

Given $\epsilon=0.01$ and $a=2$. Find a value $\delta$ corresponding to the given value of $\epsilon$ so that the definition of continuity is satisfied.

I supposed, that since $f(x)=f(a)+\epsilon$, we could solve the equation $$f(x)-f(a)=\epsilon$$$$\sqrt{x}-\sqrt{2}=0.01/-0.01$$

for $x$, choose the smallest $x$, then find the $\delta$ that is sufficiently small, so the change in the value of the function remains within $0.01$. Why doesn't this method work? The correct answer is $2\epsilon$.

Answer 1

I tried to solve it in a different way and still a dead end. Here is my work.

$$\vert \sqrt{x}-\sqrt{2} \vert<0.01$$$$-0.01<\vert \sqrt{x}-\sqrt{2}<0.01$$

$$-0.01+\sqrt{2} < \sqrt{x} < 0.01+\sqrt{x}$$

$$(-0.01+\sqrt{2})^2 < x < (0.01+\sqrt{x})^2$$

$$-0.028184 < x - 2 < 0.028384$$

The correct answer should be $\delta = 0.028184$, but the given answer is $\delta = 0.02$. Is this because any smaller $\delta$ is also a valid answer? Seems a bit arbitrary to me. This is all from the book A first course in real analysis - Protter, Morrey.

Answer 2

The function $\sqrt x$ is monotonic, so that in an interval it achieves the minimum and maximum value at the endpoints.

We try $\delta=10^{-10}$:

$$\left|\sqrt{2\pm10^{-10}}-\sqrt2\right|=\frac{10^{-10}}{\sqrt{2\pm10^{-10}}+\sqrt2}<10^{-10}<0.01$$ is true and we are done.