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How do you prove this using the epsilon-delta definition? I'm unsure of using the min = { } function.

$\lim \limits_{x \to \infty}\frac{2x+1}{1-x}$

These are my steps:

$ |f(x) - L| < \epsilon => |\frac{2x+1}{1-x} +2|< \epsilon $

$ \qquad \qquad \; \; \; \; \; =>|\frac{3}{1-x} | < \epsilon $

$ \qquad \qquad \; \; \; \; \; =>|\frac{-3}{x-1} | < \epsilon$

$ \qquad \qquad \; \; \; \; \; =>|-3||\frac{1}{x-1} | < \epsilon$

$ \qquad \qquad \; \; \; \; \; =>|\frac{1}{x-1} | < \frac{\epsilon}{3}$

$ \qquad \qquad \; \; \; \; \; =>|{x-1} | < \frac{\epsilon |x-1|}{3}$

How do I continue from here?

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  • $\begingroup$ I got stuck at |1/(1-x)|<ε, I can't figure how to achieve |x-a|<δ $\endgroup$ – G.L. May 27 '18 at 16:59
  • $\begingroup$ Achieve $|x-a| < \delta $? How are you approaching this? $\endgroup$ – JuliusL33t May 27 '18 at 17:02
  • $\begingroup$ |f(x) - L| < ε => |x-1| < ε|x-1|/3 $\endgroup$ – G.L. May 27 '18 at 17:04
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A few pointers:

  • The $\varepsilon$-$\delta$ definition of a limit pertains to when you have a limit as $x \to a$, where $a$ is a finite quantity. For a limit as $x \to \infty$, you need an $\varepsilon$-$M$ definition, along the lines of, for all $\varepsilon > 0$, there exists some $M$ such that $$x > M \implies |f(x) - L| < \varepsilon.$$
  • Your last line has an error. You'll want to be reciprocating both sides, to get $|x - 1| > \frac{3}{\varepsilon}$. Note the change in sign, as the function $\frac{1}{x}$ is decreasing for positive $x$.
  • This means, you are looking for values of $x$ that are a greater distance than $\frac{3}{\varepsilon}$ from $1$. This suggests a that $M = 1 + \frac{3}{\varepsilon}$ is a good choice of $M$. Note that no maximums or minimums are required!
  • Beware the flow of logic here! You've written in implications $\implies$ between each step. While this is true, what you really need for the definition is backwards implications: $\impliedby$. This is because you're starting with $|f(x) - L| < \varepsilon$, the thing you need to conclude! This is ok, provided each step you perform implies the previous step, not the next step as we often tend to do. The other way is to simply rewrite the entire proof backwards, starting with assuming $x > 1 + \frac{3}{\varepsilon}$, and working logically to the desired conclusion.

EDIT: Regarding the solution you posted as an answer, it's definitely looking better, though I have a couple more pointers:

  • You only really need the last block of implications. Your entire solution could be presented by simply starting with the line that begins $$|x - 1| > \frac{3}{\varepsilon} \iff \ldots$$ and going from there. Sure, it doesn't explain where you got $\frac{3}{\varepsilon}$ from, but then again, it doesn't have to! The logic shows that such a choice was a good choice, no matter how you came by it.
  • You still haven't come from the starting point of $x > M$, which you should, as this is part of the definition of a limit as $x \to \infty$. So, I would just add to your solution, \begin{align*} x > 1 + \frac{3}{\varepsilon} &\implies x - 1 > \frac{3}{\varepsilon} > 0 \\ &\implies |x - 1| > \frac{3}{\varepsilon} \\ &\implies \ldots \end{align*} Note that the above implication works because $x - 1 \ge 0$, and so $|x - 1| = x - 1$.
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Thanks, Theo Bendit. I re-did my answer, not sure if it's correct.

$ |f(x) - L| = |\frac{2x+1}{1-x} +2|< \delta $

$ \qquad \quad \; \; \; =|\frac{3}{1-x} | < \delta $

$ \qquad \quad \; \; \; =|\frac{-3}{x-1} | < \delta $

$ \qquad \quad \; \; \; =|-3||\frac{1}{x-1} | < \delta $

$ \qquad \quad \; \; \; =|\frac{1}{x-1} | < \frac{\delta}{3}$

$ \qquad \quad \; \; \; =|{x-1} | > \frac{3}{\delta}$

$ \frac{3}{\delta} = \epsilon <=> \delta = \frac{3}{\epsilon}$

$ |x-1| > \frac {3}{\epsilon} <=> 0 < |\frac{1}{x-1}|<\frac{\epsilon}{3}$

$\qquad \qquad \; \; <=> 0 < |\frac{3}{x-1}|<\epsilon$

$\qquad \qquad \; \; <=> 0 < |-3||\frac{1}{x-1}|<\epsilon$

$\qquad \qquad \; \; <=> 0 < |\frac{3}{1-x}|<\epsilon$

$\qquad \qquad \; \; <=> 0 < |\frac{2x+1}{1-x} + 2|<\epsilon$

$\qquad \qquad \; \; => 0 < |f(x) - L|<\epsilon$

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  • $\begingroup$ Ok, I'M confused looking at your answer. You're overthinking this, Theo Bendit.more or less gave you an outline of the solution. Hint: Consider that $\frac {3}{\epsilon} $ > 0 for any value of x, so you can simply drop the absolute value in | x- 1| . Then just proceed with the algebra to get his suggested value of M. $\endgroup$ – Mathemagician1234 May 28 '18 at 2:48
  • $\begingroup$ Hi G.L., I've edited my answer with a bit of feedback about your answer. Hope you find it helpful. $\endgroup$ – Theo Bendit May 28 '18 at 4:13

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