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For example: We shall prove that $$ \lim_{x \to 2} x^2 = 4.$$ Then we shall prove the statement $$ \forall \varepsilon >0, \,\, \exists \delta >0 \,\, \mid \,\, 0 < \left| x - 2\right| < \delta \implies \left|x^2 - 4 \right| < \varepsilon$$

Draft: $$ \left|x+2 \right| \left|x-2 \right| < \varepsilon \qquad (1) $$ Let $\delta \leq 1$: $$ \left|x - 2 \right| + |2| < 1 + |2|$$ $$ |(x - 2) + (2)| \leq \left|x - 2 \right| + |2| $$ $$ |x| < 1 + |2|$$ $$ x < 3 $$ So $$ |x-2||x+2| < 5|x-2| \qquad (2)$$


And here comes the doubt!


$$ 5 \left|x - 2 \right| < \varepsilon \qquad (3) $$


Proof:

Let $$\delta = \min \left(1, \, \dfrac{\varepsilon}{5} \right)$$ [The proof continues, but I don't have trouble with the remainder part.]


Why can we say that $ 5 \left|x - 2 \right| < \varepsilon $ ? If we add the inequalities (1) and (2), we won't get the inequality (3). Can anyone help me? Thank you.

Sorry for my bad English.

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  • $\begingroup$ You can't say it - you have an incomplete proof. You haven't picked $\delta$ yet. $\delta\leq 1$ is clearly not enough to work with any $\varepsilon$. $\delta$ should depend on $\varepsilon$. If you add the condition that $\delta<\epsilon/5$, you'd be done - so pick $\delta=\min(1,\varepsilon/10)$ for example. $\endgroup$ – Thomas Andrews May 13 '14 at 18:41
  • $\begingroup$ But it's not the proof; it's the draft. To start the proof, I need the inequality (3), but I cannot see how to obtain it. $\endgroup$ – user149844 May 13 '14 at 18:59
  • $\begingroup$ Then I don't understand your question. I assumed you were reading this proof somewhere, because you asked "Why can we say that...?" Which indicates you were confused about a step in an existing proof. My point is, you can't say that given what you have. And I showed you how to complete the proof by picking a specific $\delta$. $\endgroup$ – Thomas Andrews May 13 '14 at 19:23
  • $\begingroup$ And I was. But the proof's author recommends to write a draft (in order to guess a value for $\delta$), and, after that, start the proof. I'm confused with the inequality (3), since you will need it to find the relation between $\delta$ and $\varepsilon$. $\endgroup$ – user149844 May 13 '14 at 19:43
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I think your confusion comes from what is being assumed in the "draft". The draft is an attempt to find a $\delta$ depending on $\epsilon$ that will work for your function. Here's a (hopefully) more clear draft:

We want to find $\delta$ so that if $0<|x-2|<\delta$, then $|x^2 - 4| < \epsilon$. Assume $\delta \leq 1$. Then $|x+2|< 5$ as you have shown, and $|x^2 - 4| = |x-2||x+2|<5|x-2|$. So if $\delta$ is also at most $\frac{\epsilon}{5}$, then we have:

$$|x^2-4| = |x-2||x+2| < 5\left(\frac{\epsilon}{5}\right) = \epsilon$$

So we've shown that if $\delta = \min(1,\frac{\epsilon}{5})$, we have $|x^2 - 4| < \epsilon$, completing the proof.

To relate back to your original draft, (3) is not true based on (1) and (2). Rather, (3) is what we WANT to be true in order for the proof to work.

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  • $\begingroup$ I've finally understood the idea. Thank you! $\endgroup$ – user149844 May 14 '14 at 17:23
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We know that $\lim\limits_{x\rightarrow2}{x}=2 \tag{1}$(Hint: Try $\delta=\varepsilon$)

Note that $x^2-4=\left(x-2\right)\left(x-2\right)+4(x-2)\tag{2}$

Using $(1)$ we can say that there exist a $\delta_1$ such that $|x-2|<\sqrt{\frac{\varepsilon}{2}}$. Again using $1$ we can say that there exist a $\delta_2$ such that $|x-2|<\frac{\varepsilon}8$.

Now,there exist a $\delta=\min\{\delta_1,\delta_2\}$ such that $(2)$(with the help of triangular inequality) can be reduced to $$\left|x^2-4\right|<\left|\left(x-2\right)\right|\left|\left(x-2\right)\right|+\left|4(x-2)\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$

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There are several issues going on here. To answer your first question, we can't conclude $5|x-2| < \epsilon$.

Pretend I have chosen my favorite small number $\epsilon$. To find $\delta$, I need to figure out how close $x$ must be to $2$ in order to make $|x^2 -4| < \epsilon$. So from this, hopefully you can see that $\delta$ depends on $\epsilon$. The smaller $\epsilon$ is, the smaller $\delta$ will need to be. Therefore, you should not have the line "Let $\delta \leq 1$:" in your proof. Rather, you should have

"Let $\delta = $ (some expression involving $\epsilon$)"

So maybe try something like "Let $\delta = \sqrt{\epsilon}$" and see what you can come up with. (Note: this is not a correct value of $\delta$, it's just an example to get you going.)

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