Spivak, Ch. 5 Limits, Problem 3 vii: Prove a limit $l$ of a function by showing $\delta$ such that $|f(x)-l|<\epsilon\ \forall x:\ 0<|x-a|<\delta$

Question

In Spivak, Ch. 5 on Limits, problem 3 we are asked to determine the limit $l$ for the given function and value of $a$, and to prove that it is the limit by showing how to find a $\delta$ such that $|f(x)-l|<\epsilon$ for all $x$ satisfying $0<|x-a|<\delta$.

The solution for this problem is not available in the solution manual.

Consider item $vii$, which specifies the function $f(x)=\sqrt{|x|}$ with $a=0$

Case 1: $\forall \epsilon: 0<\epsilon<1$

$$\implies \epsilon^2<\epsilon<1$$ $$|x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Case 2: $\forall \epsilon: 1\leq \epsilon < \infty$

$$\epsilon\leq\epsilon^2 \implies |x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Therefore

$$\forall \epsilon>0,|x-0|<\epsilon^2\implies|\sqrt{|x|}-0|<\epsilon$$

$$\implies \lim_\limits{x\to a} \sqrt{|x|}=0$$

Is this correct?

Answer 1

There is no need to consider two separate cases.

Given $\epsilon>0$, we pick $\delta =\epsilon^2>0$, if $|x-0|<\delta =\epsilon^2$, we take square root on both sides,

$$|\sqrt{|x|}-0|< \epsilon$$