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November 4th, 2018, 05:41 AM   #11
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Quote:
 Originally Posted by raven2k7 by the way, none of those results were correct. lol

November 4th, 2018, 06:37 PM   #12
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Quote:
 Originally Posted by raven2k7 Hi I need to answer the following equation: $\displaystyle \lim_{x\to2}$ (x³ + 3x - 1) = x3 = x to the power of 3
$\displaystyle \lim_{x \to 2} = 2^3 + 3 \cdot 2 - 1$

Can you finish it now?

-Dan

 November 5th, 2018, 06:40 AM #13 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Definition $\displaystyle \lim_{x\rightarrow a}f(x)=L$ if, given $\displaystyle \epsilon$, $\displaystyle \delta$ exists such that |f(x)-L|< $\displaystyle \epsilon$ if |x-a| < $\displaystyle \delta$. Example (OP) f(x) = x$\displaystyle ^{3}$+3x-1 Prove: $\displaystyle \lim_{x\rightarrow 2}f(x)$ = 2$\displaystyle ^{3}$+3(2)-1 = 13 = L |f(x)-L| = |x$\displaystyle ^3$+3x-14| = |(x$\displaystyle ^2$+2x+7)(x-2)| Choose x to put a limit on x$\displaystyle ^2$+2x+7: Let |x-2|<3. Then 1
November 5th, 2018, 06:55 AM   #14
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Quote:
 Originally Posted by zylo Definition $\displaystyle \lim_{x\rightarrow a}f(x)=L$ if, given $\displaystyle \epsilon$, $\displaystyle \delta$ exists such that |f(x)-L|< $\displaystyle \epsilon$ if |x-a| < $\displaystyle \delta$. Example (OP) f(x) = x$\displaystyle ^{3}$+3x-1 Prove: $\displaystyle \lim_{x\rightarrow 2}f(x)$ = 2$\displaystyle ^{3}$+3(2)-1 = 13 = L |f(x)-L| = |x$\displaystyle ^3$+3x-14| = |(x$\displaystyle ^2$+2x+7)(x-2)| Choose x to put a limit on x$\displaystyle ^2$+2x+7: Let |x-2|<3. Then 1
First you choose a delta of 3 to get the statement $|x-2|<3$,
then you take $\delta\leq \epsilon/22$. Fine, but then you never demonstrated that this delta actually works.

November 5th, 2018, 07:18 AM   #15
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Quote:
 Originally Posted by Micrm@ss First you choose a delta of 3 to get the statement $|x-2|<3$, then you take $\delta\leq \epsilon/22$. Fine, but then you never demonstrated that this delta actually works.
That was a typo, it should be |x-2|<1, which is what I actually used.

November 5th, 2018, 10:34 AM   #16
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Quote:
 Originally Posted by zylo Choose x to put a limit on x$\displaystyle ^2$+2x+7
You can't choose $x$.

Also, what's wrong with putting a whole expression inside MATH tags instead of just the odd character here and there? Is it important to you that you make posts harder to read?

Last edited by v8archie; November 5th, 2018 at 11:12 AM.

November 6th, 2018, 11:13 AM   #17
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The $\displaystyle \epsilon \delta$ proof is fundamental, but I hate to lose

Quote:
 Originally Posted by zylo limit of a sum is sum of its limits limit of a product is product of its limits.
as a take-away, because it's much more practical and I don't think OP got it.

For example,

$\displaystyle \lim_{x \rightarrow \frac{\pi}{6}}(\cos ^{3}x + x^{2}\sin x)= \lim_{x \rightarrow \frac{\pi}{6}}(\cos ^{3}x)+\lim_{x \rightarrow \frac{\pi}{6}}(x^{2}\sin x)=$
$\displaystyle (\lim \cos x)^{3}+(\lim x)^{2}(\lim \sin x) = (.866)^{3} + (\frac{\pi}{6})^{2}(.5)$

The only dicy part is $\displaystyle \lim \cos x$ and $\displaystyle \lim \sin x$. The best I could come up with is derivatives are continuous so limits exist, and derivative derivation only requires $\displaystyle \lim_{\Delta x \rightarrow 0}\sin \Delta x$ and $\displaystyle \lim_{\Delta x \rightarrow 0}\cos \Delta x$ so there is no circularity.

November 6th, 2018, 07:17 PM   #18
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Quote:
 Originally Posted by zylo The only dicy part is $\displaystyle \lim \cos x$ and $\displaystyle \lim \sin x$. The best I could come up with is derivatives are continuous so limits exist, and derivative derivation only requires $\displaystyle \lim_{\Delta x \rightarrow 0}\sin \Delta x$ and $\displaystyle \lim_{\Delta x \rightarrow 0}\cos \Delta x$ so there is no circularity.
$\displaystyle \lim_{x\rightarrow a}\sin x = \sin a$ if, given $\displaystyle \epsilon$, $\displaystyle \delta$ exists st $\displaystyle |\sin x-\sin a| <\epsilon$ if $\displaystyle |x-a| < \delta$
$\displaystyle |\sin x-\sin a|=\: \biggr\rvert 2\sin \frac{x-a}{2}\cos \frac{x+a}{2}\biggr\rvert <2 \biggr\rvert \sin\frac{x-a}{2}\biggr\rvert < \epsilon$
$\displaystyle \biggr\rvert \sin\frac{x-a}{2}\biggr\rvert < \frac{\epsilon}{2}$
$\displaystyle \biggr\rvert \sin\frac{x-a}{2}\biggr\rvert <\: \biggr\rvert\frac{x-a}{2}\biggr\rvert < \frac{\epsilon}{2}$ because chord less than arc.
$\displaystyle |x-a|< \epsilon$
$\displaystyle \delta = \epsilon$

Ultimately the proof boiled down to a geometric argument because that's how sin is defined (also defined by series).

 November 8th, 2018, 04:17 AM #19 Newbie   Joined: Sep 2018 From: Spain Posts: 19 Thanks: 0 smart puppy

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