Find a $\delta$ such that $\lvert \sqrt{x} - 1 \rvert<\varepsilon$ for all $x$ satisfying $0<\lvert x-1 \rvert <\delta$

Question

Can we choose $\delta=\varepsilon$?

This is problem 3(vi) from Spivak's Calculus chapter 5. I think he's essentially asking to prove that $\lim_{x \to 1} \sqrt{x} =1$.

Spivak's proof (in the solution manual) considers the two cases $\varepsilon >1$ and $\varepsilon<1$, and arrives at $\delta=2\varepsilon-\varepsilon^2$ for $\varepsilon<1$.

But can't we just choose $\delta=\varepsilon$? as follows:

Since $x$ is real,

$\frac{1}{\sqrt x+1}\le1$

$\frac{\lvert x-1 \rvert}{\sqrt x+1}\le\lvert x-1 \rvert$

and so

$\lvert {\sqrt x-1} \rvert =\lvert (\sqrt x-1) \cdot\frac{\sqrt x+1}{\sqrt x+1} \rvert =\lvert \frac{x-1}{\sqrt x+1} \rvert = \frac{\lvert x-1\rvert}{\sqrt x+1} \le \lvert x-1 \rvert <\delta =\varepsilon$

As required?

Answer 1

You are right, except that, when you wrote ${\sqrt x-1} \cdot\frac{\sqrt x+1}{\sqrt x+1}$, you should have written $\left({\sqrt x-1}\right)\cdot\frac{\sqrt x+1}{\sqrt x+1}$.