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 January 7th, 2019, 11:14 AM #11 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 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: 7,923 Thanks: 1123 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: 7,623 Thanks: 2611 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: 7,623 Thanks: 2611 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: 124 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.

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