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If you want to prove $\lim_{x\to2}(x^2-3x+1) = -1$

Then let $\epsilon \gt 0$ be given and we want to find a $\delta \gt 0$ such that whenever $0 \lt|x-2| \lt \delta$ we have $|(x^2-3x+1)+1| \lt \epsilon$

The general method i follow before i actually start the proof is to work backwards from that inequality to find an equivalent inequality showing the range of $x$ around $2$ which satisfies it in terms of $\epsilon$; i.e something like $|x-2| \lt \epsilon/3$ for example.

But with this example you end up with

$|(x-1)||(x-2)|<\epsilon$

and have to restrict $\delta$ as the whole point is to get $\delta$ in terms of $\epsilon$. If you restrict $\delta$ to be say $\delta \lt 1$ then i get how this restricts range of $x: 1 \lt x \lt 3$ and i have looked at the solution where they find the upper bound of $|x-1|$ to be $2$. Sub that in and then show the implication but i get lost during this whole part because i can't follow what is going on.

Could someone help me out, thanks.

Robert Shore
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Carlos Bacca
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    This seems like a perfectly reasonable question and I don't understand the downvote. The point is that in many of these proofs we need two separate constraints on $\delta$, and asking why that is and how the fixed constraint contributes to the proof seems entirely reasonable. – Robert Shore May 29 '19 at 00:26
  • You may have a look at https://math.stackexchange.com/a/2294403/72031 – Paramanand Singh May 29 '19 at 03:08

2 Answers2

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Here is a more general consideration which gives hopefully some insight. You might play it through with your specific example to get a grasp of it:

You want to prove $\lim_{x\to x_0}f(x) = L$ using $\epsilon-\delta$.

So, you start with $|f(x) - L|$ and try to bound it as follows:

  • $|f(x)-L| \leq \color{blue}{|p(x)|}|x-x_0|$, where $\color{blue}{p(x)}$ is a function defined in a neighbourhood of $x_0$ (maybe after removing $x_0$).

You want to have

  • $\color{blue}{|p(x)|}|x-x_0| \stackrel{!}{<} \epsilon$

So, you need to control $\color{blue}{|p(x)|}$. Here is the point where the restrictions for $\delta$ come into play. It is often possible to find a bound for $\color{blue}{|p(x)|}$ in a $\color{blue}{\mbox{neighbourhood}}$ of $x_0$:

  • $\color{blue}{|p(x)| \leq M}$ for $0 < |x-x_0| < \color{blue}{\delta_0}$

Now you get

$$|f(x)-L| \leq \color{blue}{|p(x)|}|x-x_0| \leq \color{blue}{M}|x-x_0| < \epsilon \mbox{ for } \delta < \min\left(\frac{\epsilon}{\color{blue}{M}} ,\color{blue}{\delta_0} \right)$$

trancelocation
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You have to use the definition. Let $\epsilon >0$, you want to find $\delta = \delta(\epsilon)>0$ (which depends of $\epsilon$) such that $$ 0<|x-2|<\delta \Rightarrow |x^2 -3x + 1 + 1| < \epsilon.$$

First you do $$ |x^2 - 3x + 1+1|=|x^2-3x + 2| = |(x-2)(x-1)|=|x-2|\cdot|x-1| $$

Then let $\delta = 1$ $$ |x-2|<\delta \Rightarrow |x-2|<1 \Rightarrow -1<x-2<1 \Rightarrow1<x<3 \Rightarrow 0<x-1<2 \Rightarrow|x-1|<2 $$

So that $$ |x-2|\cdot|x-1| < 2 \cdot |x-2| < 2 \delta = \epsilon \iff \delta = \frac{\epsilon}{2} $$

Then, you have to choose $\delta$ = min{1, $\frac{\epsilon}{2} \} $

hugdelcur96
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  • I think i understand it a bit better. So the whole point of putting a restriction on d is so we can relate |x-1||x-2|<2|x-2|<e. to do this we used the assumption that x is within 1 of 2 and so when you go the other way to prove it its only going to be a valid argument when e/2 is within 1. then we just take delta equals 1 to deal with all other e values. I think its something along these lines thanks for the answer – Carlos Bacca May 29 '19 at 01:39