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Prove $\lim_{x\to 1} {2+4x \over 3} = 2$ using the epsilon delta definition of a limit.

if $0 < \left|x-1\right| < \delta$ then $\left|{2+4x \over 3}-2\right| < \epsilon$

scratch work for finding $\delta$ $$ \left|{2+4x \over 3}-2\right| < \epsilon $$ $$ 2-\epsilon < {2+4x \over 3} < 2 + \epsilon $$ $$ 6-3\epsilon < 2+4x < 6 + 3\epsilon $$

$$ 4-3\epsilon < 4x < 4 + 3\epsilon $$

$$ 1-\frac34\epsilon < x < 1 + \frac34\epsilon $$

$$ \frac34\epsilon < x-1 < \frac34\epsilon $$

$$ \left|x-1\right| < \frac34\epsilon $$ This is where I need help.

Given $\epsilon > 0$ and let $\delta = \frac34\epsilon$

$$\left|{2+4x\over 3}-2\right| < \frac43\delta$$ $$-\frac43\delta < {2+4x \over 3}-2 < \frac43\delta $$ $$2-\frac43\delta < {2+4x\over 3}< 2 + \frac43\delta$$ $$6-4\delta<2+4x < 6+4\delta $$ $$4-4\delta<4x<4+4\delta$$ $$1-\delta<x<1+\delta$$ $$\left|x-1\right|<\delta$$

I'm pretty sure the $\frac43\delta$ part is wrong on the first line which makes the rest wrong but I'm not sure how to do this. Thanks.

alan
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2 Answers2

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Your first part of your solution is correct. You have indeed found a $\delta=\frac34\epsilon$ that will work. Then you proceed like this. $$\forall\epsilon>0,\ \exists\delta>0\ \text{such that }|x-1|<\delta\implies\left|\frac{2+4x}{3}-2\right|<\epsilon$$ Let $\delta=\frac34\epsilon$. Then, $$|x-1|<\delta\implies|x-1|<\frac34\epsilon\implies\frac43|x-1|<\epsilon\implies\left|\frac{4x-4}{3}\right|<\epsilon\quad\implies\left|\frac{2+4x-6}{3}\right|<\epsilon\implies\left|\frac{2+4x}{3}-2\right|<\epsilon\implies|f(x)-L|<\epsilon$$ And we are done.

homegrown
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  • You stated the first part of my solution is correct, and I was pretty sure it was already. What I am confused on though is how my second part is correct as stated by Lord Soth. How can I pick a constant such as $\frac43$ for the purpose of canceling out the $\frac34$ for the $\epsilon$? – alan May 05 '14 at 03:29
  • What do you mean pick a constant such as $\frac43.$ You already found a constant $\delta=\frac34\epsilon.$ Now, the $\frac34$ cancelled out from the $\epsilon$ by multiplying each side of the inequality by the reciprocal, $\frac43$. See my proof. Then I just did manipulations to arrive at the $|f(x)-L|<\epsilon$, which was what we wanted to show. – homegrown May 05 '14 at 04:08
  • ok, i see what you did – alan May 05 '14 at 04:21
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Nothing is wrong with what you have done. For every $\epsilon>0$, you needed to find a $\delta > 0$ such that $0<|x-1|<\delta \implies |\frac{2+4x}{3}-2|<\epsilon$. Your first derivations (before you say "This is where I need help") actually find such a $\delta$ (as they are all "if and only if" statements). That is, you have shown $\delta = \frac{3}{4}\epsilon$ works. I suggest you take some time for your proof to "sink in" so as to clear any confusion that you may have.

Lord Soth
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  • You stated the first part of my solution is correct. What I am confused on though is how my second part is correct. How can I pick a constant such as $\frac43$ for the purpose of canceling out the $\frac34$ for the ϵ? Or I guess, what does that mean when I do that? I am just following an example from a book without understanding it, but I do not know how to understand it. – alan May 05 '14 at 03:36