November 4th, 2018, 05:41 AM  #11 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552  
November 4th, 2018, 06:37 PM  #12 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,340 Thanks: 983 Math Focus: Wibbly wobbly timeywimey stuff.  
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 xa < $\displaystyle \delta$. Example (OP) f(x) = x$\displaystyle ^{3}$+3x1 Prove: $\displaystyle \lim_{x\rightarrow 2}f(x)$ = 2$\displaystyle ^{3}$+3(2)1 = 13 = L f(x)L = x$\displaystyle ^3$+3x14 = (x$\displaystyle ^2$+2x+7)(x2) Choose x to put a limit on x$\displaystyle ^2$+2x+7: Let x2<3. Then 1<x<3 and 10<x$\displaystyle ^2$+2x+7<22.. Then f(x)L<22x  2<$\displaystyle \epsilon$ if x2< $\displaystyle \epsilon$/22. So f(x)L<$\displaystyle \epsilon$ if $\displaystyle \delta \leq 1$ and $\displaystyle \delta \leq \epsilon/22$ 
November 5th, 2018, 06:55 AM  #14  
Senior Member Joined: Oct 2009 Posts: 905 Thanks: 354  Quote:
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 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126  
November 5th, 2018, 10:34 AM  #16 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra  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  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
The $\displaystyle \epsilon \delta$ proof is fundamental, but I hate to lose Quote:
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  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126  Quote:
$\displaystyle \sin x\sin a=\: \biggr\rvert 2\sin \frac{xa}{2}\cos \frac{x+a}{2}\biggr\rvert <2 \biggr\rvert \sin\frac{xa}{2}\biggr\rvert < \epsilon$ $\displaystyle \biggr\rvert \sin\frac{xa}{2}\biggr\rvert < \frac{\epsilon}{2}$ $\displaystyle \biggr\rvert \sin\frac{xa}{2}\biggr\rvert <\: \biggr\rvert\frac{xa}{2}\biggr\rvert < \frac{\epsilon}{2}$ because chord less than arc. $\displaystyle xa< \epsilon$ $\displaystyle \delta = \epsilon$ 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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