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January 5th, 2019, 05:26 AM   #1
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limit

Using delta- epsilon definition, please give me a proof of how limit x tends to infinity of the function f(x) = 1/x is not equal to 1.

I am confused in the choice of delta.
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January 5th, 2019, 01:14 PM   #2
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You can do the mechanics. $\lim_{x\to \infty} |\frac{1}{x}-1|=1\ne 0$.
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January 5th, 2019, 05:33 PM   #3
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The definition of the limit says that $$\lim_{x \to \infty} \frac1x = 1$$
means that
Quote:
for every $\epsilon > 0$ there exists an $x_0$ such that $\left|\frac1x - 1\right| < \epsilon$ for every $x > x_0$.
You have to prove that statement to be false. This entails finding a value of $\epsilon$ for which no such $x_0$ exists - any $x_0$ you pick has some value of $x > x_0$ such that $\left|\frac1x - 1\right| \not < \epsilon$.

I suggest that you consider some value of $\epsilon < 1$ - for example $\epsilon = \frac12$.

Last edited by v8archie; January 5th, 2019 at 05:41 PM.
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January 6th, 2019, 06:38 AM   #4
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Alternatively, you could prove that the limit is something other than 1.
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January 6th, 2019, 06:42 AM   #5
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Alternatively, you could prove that the limit is something other than 1.
You'd still need to prove that the limit is unique.
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January 6th, 2019, 09:10 AM   #6
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You'd still need to prove that the limit is unique.
Ahh, yes.
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January 6th, 2019, 04:06 PM   #7
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Originally Posted by Micrm@ss View Post
You'd still need to prove that the limit is unique.
If there is a limit, it must be unique(??)
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January 6th, 2019, 07:26 PM   #8
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If there is a limit, it must be unique(??)
Sure, but that needs to be proven.
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January 7th, 2019, 10:42 AM   #9
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Sure, but that needs to be proven.
Of course, it could be proved generally as a theorem and then applied to specific problems, but I must admit that I merely assumed the general theorem and neglected to prove it.
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January 7th, 2019, 10:57 AM   #10
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The derivative of f(x) = 1/x, -1/x$^2$, tells us f(x) is monotone decreasing. Is that sufficient to tell us the limit is unique (if it exits)? I think it is.
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