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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 $f(x)=\frac{1-\sqrt{x}}{1-x}$:

Graph

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:

  1. Is this a valid technique for epsilon-delta proofs?
  2. Are there simpler ways of approaching this problem?
Joe
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    Factor the denominator. That will make life simpler. – Cameron L. Williams Mar 13 '21 at 14:05
  • It just makes things more complicated and harder for you. Instead of finding any $\delta$ which works you are putting unnecessary effort to find a very special $\delta$. We are not even guaranteed that a maximum $\delta$ would exist for a given limit problem. – Paramanand Singh Mar 13 '21 at 16:06
  • Also solving inequalities to find $\delta$ as an explicit expression in terms of $\epsilon $ is against the whole spirit of limit definition. You need to ditch the attitude of algebraic manipulation of inequalities and instead focus on how to find a very gross estimate for upper bound of $|f(x) - L|$ in a simple form $K|x-a|$ – Paramanand Singh Mar 13 '21 at 16:09
  • The fact that we don't need any sharp estimates here gives a lot of room / leeway and makes our work considerably simpler. An analogy: finding an approximation for $\sqrt{2}$ valid upto one decimal place is far easier than finding another one which is valid upto five decimal places. The limit definition is far simpler to apply than typical textbook exercises will make you believe. – Paramanand Singh Mar 13 '21 at 16:13
  • @ParamanandSingh: What strategy would you suggest that I use instead? – Joe Mar 13 '21 at 19:14
  • See in particular this answer. Also note that when dealing with limit of $f(x) $ via $\epsilon,\delta$ one can't use the inverse function $f^{-1}$ to get expression for $\delta$ in terms of $\epsilon $. – Paramanand Singh Mar 14 '21 at 01:53

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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 $

sirous
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    The point of the question was to give feedback on the technique used in $\epsilon, \delta$ proof. The asker is already aware of typical methods to evaluate the limit. – Paramanand Singh Mar 13 '21 at 16:17