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 January 7th, 2019, 11:14 AM #11 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra How are you thinking? If the two limits are $a$ and $b$ such that $0 \ge a > b$ and $f(x) \ge a$ for all values of $x$, how does the fact that $f(x)$ is monotonically decreasing lead to a contradiction? January 7th, 2019, 02:27 PM #12 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond If f(x) $\ge$ a how can we say f(x) approaches b? January 7th, 2019, 04:19 PM #13 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra But that argument doesn't use the monotone property of $f(x)$, it falls back to the $\epsilon-\delta$ definition of the limit, doesn't it? January 7th, 2019, 04:49 PM   #14
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 Originally Posted by v8archie But that argument doesn't use the monotone property of $f(x)$...
I think it does as I am interpreting monotonicity as implying a function steadily and evenly approaches its limit. January 7th, 2019, 07:47 PM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra But that sentence itself implicitly states that there is only one limit. January 7th, 2019, 07:48 PM   #16
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 Originally Posted by v8archie How are you thinking? If the two limits are $a$ and $b$ such that $0 \ge a > b$ and $f(x) \ge a$ for all values of $x$, ...  January 11th, 2019, 06:07 AM #17 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Definition: $\displaystyle \lim_{x\rightarrow \infty}\frac{1}{x} = L$ if, given $\displaystyle \epsilon$, M exists such that $\displaystyle \left|\frac{1}{x}-L\right|<\epsilon$ if $x$ > M Let $\displaystyle \epsilon = .01$ $\displaystyle \left|\frac{1}{x}-1\right|<.01$ $\displaystyle -.01<\frac{1}{x}-1<.01 \rightarrow$ $\displaystyle .99<\frac{1}{x}<1.01$ There is no M such that this is true for all $x$ > M, so $L$ can't be 1. Last edited by skipjack; January 18th, 2019 at 06:20 AM. Tags limit 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 zylo Calculus 13 May 31st, 2017 12:53 PM veronicak5678 Real Analysis 4 August 22nd, 2011 10:07 AM unlimited Algebra 7 April 3rd, 2011 10:14 AM madzia Algebra 0 January 23rd, 2010 01:52 PM conjecture Calculus 1 July 24th, 2008 01:14 PM

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