Is being 'economical' a good strategy for epsilon-delta proofs? Today I was trying to compute the following limit:
$$
\DeclareMathOperator{\epsilon}{\varepsilon}
\lim_{x \to 1}\frac{1-\sqrt{x}}{1-x}
$$
A single application of L'Hôpital's rule tells us that the limit is equal to $1/2$, but I wanted to practise proving that it is equal to $1/2$ directly using the epsilon-delta definition of a limit. The strategy that I used seems somewhat unorthodox, since I have never seen it in a calculus textbook. My idea was to find the maximum value of $\delta$ for which it is still true that if $x\in(1-\delta,1+\delta)$ then $\frac{1-\sqrt{x}}{1-x}\in(1/2-\epsilon,1+\epsilon)$, provided that $x\neq1$. Consider the graph of $f(x)=\frac{1-\sqrt{x}}{1-x}$:

If $x\in[1-\delta,1+\delta]$, then the value of $x$ for which $f(x)$ is furthest away from $1/2$ is $1-\delta$ (this is because the gradient of $f$ is strictly increasing). Hence, if $f(1-\delta)-1/2=\epsilon$, then it must be the case that if $x\in(1-\delta,1+\delta)$, then $\frac{1-\sqrt{x}}{1-x}\in(1/2-\epsilon,1+\epsilon)$. All we have to do is solve $f(1-\delta)-1/2=\epsilon$:
$$
\frac{1-\sqrt{1-\delta}}{\delta}-\frac{1}{2} =\epsilon \, .
$$
After some algebraic manipulation, we find that
$$
\delta = \frac{8\epsilon}{(2\epsilon+1)^2} \, ,
$$
which completes the proof. My questions are:

*

*Is this a valid technique for epsilon-delta proofs?

*Are there simpler ways of approaching this problem?


Simpler way is:
$$\frac{1-\sqrt x}{1-x}=\frac{1-\sqrt x}{(1-\sqrt x)(1+\sqrt x)}=\frac 1{1+\sqrt x}$$
$\Rightarrow Lim_{x\rightarrow 1}\frac{1-\sqrt x}{1-x}=\frac 12 $
