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So I have to prove $\lim\limits_{x \to 2} \frac{x-1}{x^2}$= $\frac{1}{4}$.

So I need to find $0<|x-2|<\delta$, such that $\left\vert\frac{x-1}{x^2}-\frac{1}{4}\right\vert<\epsilon$.

So $\left\vert\frac{x-1}{x^2}-\frac{1}{4}\right\vert=\frac{(x-2)^2}{4x^2}$.

Now I try to set $\delta=1$, so $0<|x-2|<1$ and $4<4x^2<36$.

And so $\frac{(x-2)^2}{4x^2}<\frac{(x-2)^2}{4}$.

And so $\frac{(x-2)^2}{4}<\epsilon$, when $\delta=\sqrt{4\epsilon}$.

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  • $\begingroup$ How did you arrive at the lower bound of $0$ in $\color{red}{0}<4x^2<36$? If $|x-2| < 1$, then $1 < x < 3$ so $4 < 4x^2 < 36$. $\endgroup$ – StackTD Jun 10 '16 at 11:54
  • $\begingroup$ Edited. I was one step ahead of myself, thinking 4*0=0. I forgot to add across both sides first, $\endgroup$ – stackdsewew Jun 10 '16 at 11:57
  • $\begingroup$ Wait so I can just set delta to be square root of 4*epsilon, right? $\endgroup$ – stackdsewew Jun 10 '16 at 11:59
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    $\begingroup$ You use it in the step where you replace the denominator $4x^2$ by the (lower bound, to get an upper bound on the fraction) $4$. $\endgroup$ – StackTD Jun 10 '16 at 12:04
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    $\begingroup$ @anna_xox You can just set $\delta$ to be $2 \sqrt{\epsilon}$ since inequality is carried on from the previous steps. Since you also want $\delta = 1$, we take the minimum. That is, $\delta = \text{min} \{ 1, 2 \sqrt{\epsilon}\}$. Do you understand why we take the minimum? $\endgroup$ – Jester Tran Jun 10 '16 at 12:09
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Looks good, because you can use $\color{blue}{4<4x^2}<36$ to say: $$|f(x)-L|=\frac{(x-2)^2}{4x^2} \le \frac{(x-2)^2}{4}$$ And since $|x-2| < \delta$, this means: $$\frac{(x-2)^2}{4} < \frac{\delta^2}{4}$$ You want this under $\varepsilon$, so: $$\frac{\delta^2}{4} < \varepsilon \iff \delta < 2\sqrt{\varepsilon}$$ Note that you know require $\delta \le \min\left\{ 1,2\sqrt{\varepsilon} \right\}$.


Coming back to the '<' vs '=' (see comments as well): if you set $\delta = 1$ along the way and you end up with requiring $\delta = 2\sqrt{\varepsilon}$, note that you need the strongest (i.e. smallest) bound on $\delta$.

If you have a good $\delta$, any $\delta' < \delta$ will work as well; which is why you'll often see $\delta$ taken to be smaller than (or equal to) all the upper bounds you set / need in your proof.

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May be you set $\delta$ to 1,and $0<|x-2|<1$, so $ 4 x^2 $ will between 4 and 36, that is $ 1/4 x^2<1/4 $. Then we can get $ (x-2)^2/(4 x^2)< \frac{1}{4} (x-2)^2 $. Set $ \frac{1}{4} (x-2)^2 < \epsilon $, then $ |x-2|<2 \sqrt{ \epsilon} $. Let $ \delta = Min \{ 2\sqrt \epsilon ,1 \} $

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  • $\begingroup$ I think your reasoning is circular. We are trying to find a delta in terms of an epsilon. You have found a delta in terms of a delta. $\endgroup$ – stackdsewew Jun 10 '16 at 12:06
  • $\begingroup$ Oh~It's my first edition.I can explain it. Set $1/16 \delta^2<\epsilon$,get the $\delta$,can solve it. $\endgroup$ – user346926 Jun 10 '16 at 12:12
  • $\begingroup$ Your 1/16 should be 1/4. $\endgroup$ – stackdsewew Jun 10 '16 at 12:13
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    $\begingroup$ I correct it as you say. I also a beginner learn Calculus and $ \LaTeX $. $\endgroup$ – user346926 Jun 10 '16 at 12:24

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