January 7th, 2019, 12:14 PM  #11 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 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, 03:27 PM  #12 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond 
If f(x) $\ge$ a how can we say f(x) approaches b?

January 7th, 2019, 05:19 PM  #13 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 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, 05:49 PM  #14 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond  
January 7th, 2019, 08:47 PM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 Math Focus: Mainly analysis and algebra 
But that sentence itself implicitly states that there is only one limit.

January 7th, 2019, 08:48 PM  #16 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond  
January 11th, 2019, 07:07 AM  #17 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,641 Thanks: 119 
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 07:20 AM. 

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