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$\lim_{x \to 27} x^{1/3}=(27^{1/3})$

let $x$ be $0<|x-27|< \delta$ What I have done is: $|x^{1/3}-27^{1/3}|= |x-27|/|x^{2/3}+3x^{1/3}+9|$

And I was given the solution that sets $\delta=min(26,13\epsilon)$ I understand what it means but What I want to know is: is there any standard approach for solving this kind of question? like why we guess delta=26 in the first place?

thanks!

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  • $\begingroup$ I think that you should try $0<x-27<\delta$ and $0<27-x<\delta$ separately. It will make your derivation easier. $\endgroup$ – mike Nov 30 '14 at 6:42
  • $\begingroup$ @mike Now I have this : 0<|x-27|<26 given in my answer, but Im not sure why delta set to be 26 in the first place $\endgroup$ – UnusualSkill Nov 30 '14 at 6:43
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It is not important what specific value you choose: You have only one goal:

$$|x^{1/3} - 27^{1/3}| < \epsilon .$$

And you choose $\delta$ so small to make this happen.

You know already that $$|x^{1/3} - 27^{1/3}| = \frac{|x-27|}{x^{2/3} + 3x^{1/3} + 9}.$$

Note that you what to make a lower bound $C$ on $x$, so that

$$(*) \frac{|x-27|}{x^{2/3} + 3x^{1/3} + 9} \leq \frac{|x-27|}{C^{2/3} + 3C^{1/3} + 9}.$$

Then you want to choose a $C$ so that the term is easy to compute, so let's choose $C=1$. Which means that $x\geq 1$, and that can be done if you choose $|x-27| \leq 26$.

So you have

$$\frac{|x-27|}{x^{2/3} + 3x^{1/3} + 9} \leq \frac{|x-27|}{1 + 3 + 9} = \frac{|x-27|}{13}. $$

Now it's easy to make the RHS $\leq \epsilon$: Choose $\delta < 13\epsilon$.

So you see, the choice of that 26 only help you calculate $C^{2/3} + 3C^{1/3} + 9$ easier. You can as well choose another constant. For example, I choose $C = 0$. That is, $x\geq 0$. So I need that $|x-27|\leq 27$. Then if this is true, we have

$$\frac{|x-27|}{x^{2/3} + 3x^{1/3} + 9} \leq \frac{|x-27|}{9}. $$

So you see that we can also choose $\delta = \min\{ 27, 9\epsilon\}$.

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  • $\begingroup$ I don quite understand why when choose C=1, then means X>=1? $\endgroup$ – UnusualSkill Nov 30 '14 at 7:04
  • $\begingroup$ We want to choose $C$ a lower bound for $x$, that is $x\geq C$. (So that (*) is true) @UnusualSkill $\endgroup$ – user99914 Nov 30 '14 at 7:06
  • $\begingroup$ can I say like this? only when x>=1,then this inequality is true |x^2/3+3x^1/3+9|>1+3+9 $\endgroup$ – UnusualSkill Nov 30 '14 at 7:08
  • $\begingroup$ @UnusualSkill: Yes, that's the same. $\endgroup$ – user99914 Nov 30 '14 at 7:09
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    $\begingroup$ If $C = 0$, that main we choose $x\geq 0$, not $\leq 0$. @UnusualSkill $\endgroup$ – user99914 Nov 30 '14 at 7:23

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