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January 7th, 2019, 11:14 AM   #11
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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?
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January 7th, 2019, 02:27 PM   #12
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If f(x) $\ge$ a how can we say f(x) approaches b?
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January 7th, 2019, 04:19 PM   #13
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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?
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January 7th, 2019, 04:49 PM   #14
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Originally Posted by v8archie View Post
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.
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January 7th, 2019, 07:47 PM   #15
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But that sentence itself implicitly states that there is only one limit.
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January 7th, 2019, 07:48 PM   #16
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Originally Posted by v8archie View Post
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 $x$, ...
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January 11th, 2019, 06:07 AM   #17
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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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