Help with using the formal definition of limit to show that the limit [x -> 1/4] [8x] = 2
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Dr.Peterson
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They did that because the goal is to find a set of values of x near 1/4 within which it is known that 8x will be "near" 2. They can't stop with the latter fact (epsilon), because the definition of a limit requires finding a "delta" that works.
Or are you asking a different kind of "why" question? A little explanation of your thinking might help communicate better.
Or are you asking a different kind of "why" question? A little explanation of your thinking might help communicate better.
Dr.Peterson
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Delta can't just be epsilon because then things wouldn't work.
You want to be able to prove that whenever [MATH]\left|x - \frac{1}{4}\right| < \delta[/MATH], then [MATH]\left|8x - 2\right| < \epsilon[/MATH]. But if you took delta to be epsilon, then if [MATH]\left|x - \frac{1}{4}\right| < \epsilon[/MATH], it would only follow that [MATH]\left|8x - 2\right| = \left|8\left(x - \frac{1}{4}\right)\right| = 8\left|x - \frac{1}{4}\right| < 8\epsilon[/MATH]. That lets it be too big.
I think introductory examples of these proofs may too often omit this step of stating clearly just what has been shown, because it is obvious to us who write them! The process we show is how to find a delta that will work, and hidden in that process is the fact that it does in fact work. But the solution you show does exactly this, in the last three lines. They are showing there that this choice of delta does what we need it to do. (On the other hand, you could choose any value of delta less than epsilon/8, and it would still work. We only need to find a delta that works, not the only, or the best, delta.)
By the way, you should try to type in what you can rather than just use images, to help others quote you and interact more easily. When there are special characters, you can either paste them in from some source such as Windows Character Map, or use the calculator icon in the tool bar to surround "\delta" or "\epsilon", which is the simplest use of the LaTeX formatting system. You don't need to do all the fancy stuff I did above, but a little bit helps a lot.
You want to be able to prove that whenever [MATH]\left|x - \frac{1}{4}\right| < \delta[/MATH], then [MATH]\left|8x - 2\right| < \epsilon[/MATH]. But if you took delta to be epsilon, then if [MATH]\left|x - \frac{1}{4}\right| < \epsilon[/MATH], it would only follow that [MATH]\left|8x - 2\right| = \left|8\left(x - \frac{1}{4}\right)\right| = 8\left|x - \frac{1}{4}\right| < 8\epsilon[/MATH]. That lets it be too big.
I think introductory examples of these proofs may too often omit this step of stating clearly just what has been shown, because it is obvious to us who write them! The process we show is how to find a delta that will work, and hidden in that process is the fact that it does in fact work. But the solution you show does exactly this, in the last three lines. They are showing there that this choice of delta does what we need it to do. (On the other hand, you could choose any value of delta less than epsilon/8, and it would still work. We only need to find a delta that works, not the only, or the best, delta.)
By the way, you should try to type in what you can rather than just use images, to help others quote you and interact more easily. When there are special characters, you can either paste them in from some source such as Windows Character Map, or use the calculator icon in the tool bar to surround "\delta" or "\epsilon", which is the simplest use of the LaTeX formatting system. You don't need to do all the fancy stuff I did above, but a little bit helps a lot.
pka
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The problem is \(\displaystyle \mathop {\lim }\limits_{x \to \frac{1}{4}} \left( {8x} \right) = 2\)
Whoever wrote that non-proof tried to show the reasoning in reverse.
We need to show that it possible to specify a \(\displaystyle \delta\)-neighborhood, \(\displaystyle |x-\tfrac{1}{4}|<\delta\) so that for any value of \(\displaystyle x\in(\tfrac{1}{4}-\delta,\tfrac{1}{4}+\delta\) then it must be true that \(\displaystyle |8x-2|<\varepsilon\) for a previously given \(\displaystyle \varepsilon>0\)
Here is a story you are working with someone else who says to you, "we need to be sure that we are within \(\displaystyle .07\) of \(\displaystyle 2\) when \(\displaystyle x\) ts multiplied by \(\displaystyle 8\)."
You see that you are asked to control the size of \(\displaystyle \delta\) but you are given \(\displaystyle \varepsilon\) and cannot change it.
So let's start with \(\displaystyle \delta=1\), we are free to do that.
\(\displaystyle \left| {x - \tfrac{1}{4}} \right| < 1 \)
\(\displaystyle -1+\tfrac{1}{4}<x<1+\tfrac{1}{4}\)
\(\displaystyle -\tfrac{3}{4}<x<\tfrac{5}{4}\)
\(\displaystyle -6<8x<10\)
\(\displaystyle -8<8x-2<8\)
\(\displaystyle |8x-2|<8\)
Because you are free to chose \(\displaystyle \delta\) let \(\displaystyle \delta=\min\left\{1,\tfrac{\varepsilon}{8}\right\}\)
Now you have If \(\displaystyle \left| {x - \tfrac{1}{4}} \right| < \delta \) then \(\displaystyle \left| 8x - 2\right| < \varepsilon\).
That is showing what must be done.
Whoever wrote that non-proof tried to show the reasoning in reverse.
We need to show that it possible to specify a \(\displaystyle \delta\)-neighborhood, \(\displaystyle |x-\tfrac{1}{4}|<\delta\) so that for any value of \(\displaystyle x\in(\tfrac{1}{4}-\delta,\tfrac{1}{4}+\delta\) then it must be true that \(\displaystyle |8x-2|<\varepsilon\) for a previously given \(\displaystyle \varepsilon>0\)
Here is a story you are working with someone else who says to you, "we need to be sure that we are within \(\displaystyle .07\) of \(\displaystyle 2\) when \(\displaystyle x\) ts multiplied by \(\displaystyle 8\)."
You see that you are asked to control the size of \(\displaystyle \delta\) but you are given \(\displaystyle \varepsilon\) and cannot change it.
So let's start with \(\displaystyle \delta=1\), we are free to do that.
\(\displaystyle \left| {x - \tfrac{1}{4}} \right| < 1 \)
\(\displaystyle -1+\tfrac{1}{4}<x<1+\tfrac{1}{4}\)
\(\displaystyle -\tfrac{3}{4}<x<\tfrac{5}{4}\)
\(\displaystyle -6<8x<10\)
\(\displaystyle -8<8x-2<8\)
\(\displaystyle |8x-2|<8\)
Because you are free to chose \(\displaystyle \delta\) let \(\displaystyle \delta=\min\left\{1,\tfrac{\varepsilon}{8}\right\}\)
Now you have If \(\displaystyle \left| {x - \tfrac{1}{4}} \right| < \delta \) then \(\displaystyle \left| 8x - 2\right| < \varepsilon\).
That is showing what must be done.