Calculus Calculus Math Forum

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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 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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 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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 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 Tags limits Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post consigliere- Calculus 4 January 30th, 2013 06:54 AM math89 Calculus 2 January 26th, 2013 05:11 PM Arley Calculus 2 April 2nd, 2012 06:50 PM kadmany Calculus 9 March 18th, 2011 07:16 AM lilwayne Calculus 18 September 23rd, 2010 03:39 PM

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