Evaluating limits

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Homework Statement



Consider a function f: D∈R, where D = {1/n for natural numbers n (1, 2, 3, 4, etc.)} and f(x) = 3x - 1 for all x in D. Explain why the limit of f(x) as x → 1 does not exist.

Homework Equations





The Attempt at a Solution



Uh I figured it would exist. We know a function does not have a limit at c if and only if there exists a sequence (x_n) where x_n ≠ c for all natural numbers n such that (x_n) converges to c but the sequence (f(x_n)) does not converge in the reals.

there will not exist such a sequence in D that converges to 1 because x_n cannot equal 1 for any n. so the limit must exist. i mean why wouldn't the limit exist?
 

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  • #2
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Homework Statement



Consider a function f: D∈R, where D = {1/n for natural numbers n (1, 2, 3, 4, etc.)} and f(x) = 3x - 1 for all x in D. Explain why the limit of f(x) as x → 1 does not exist.

Homework Equations





The Attempt at a Solution



Uh I figured it would exist. We know a function does not have a limit at c if and only if there exists a sequence (x_n) where x_n ≠ c for all natural numbers n such that (x_n) converges to c but the sequence (f(x_n)) does not converge in the reals.

there will not exist such a sequence in D that converges to 1 because x_n cannot equal 1 for any n. so the limit must exist.
x1 = 1/1 = 1, right?
i mean why wouldn't the limit exist?
 
  • #3
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x1 = 1/1 = 1, right?

does this have to do with the fact that the limit from the left does not exist? but 1/n is always less than or equal to 1...so things seem well-defined....
 
  • #4
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The limit from the left has a better chance of existing than the limit from the right, which is not to say that the limit from the left exists. It might be helpful to sketch a graph of f, keeping in mind that the inputs to f come from your set D.
 
  • #5
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i don't quite understand the notion of a limit from one direction having a better chance of existing. to me, it is extremely clear the limit from the left does not exist, so the limit itself doesn't exist.

but things are confusing because we're using 1/n, which goes a different direction than x. if that makes sense.
 
  • #6
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By "better chance" I wasn't implying that the limit from the left actually existed. What I was getting at is that there are numbers in D that are smaller than 1, but no numbers in D that are larger than 1, so there's no chance of a limit from the right existing.

stripes said:
but things are confusing because we're using 1/n, which goes a different direction than x. if that makes sense.
It might be helpful to look at set D like this:
D = {..., 1/5, 1/4, 1/3, 1/2, 1}
Note that you can never get closer to 1 than 1/2 for the numbers in D.
 

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