Limit of a piece-wise sequence

[wolf441]

What is the limit for this sequence?

\(\displaystyle (f)\, a_n\, =\, \begin{cases} e^n, &\mbox{ for }\, 0\, \leq\, n\, < 100 \\ e^{-n}, &\mbox{ for }\, n\, \geq\, 100 \end{cases}\)

the answer says its 0. But, I'm convinced the limit doesn't exist.

 

[Deleted member 4993]

What is the limit for this sequence?

\(\displaystyle (f)\, a_n\, =\, \begin{cases} e^n, &\mbox{ for }\, 0\, \leq\, n\, < 100 \\ e^{-n}, &\mbox{ for }\, n\, \geq\, 100 \end{cases}\)

the answer says its 0. But, I'm convinced the limit doesn't exist.

Why do you think limit does not exist at n → ∞?

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

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http://www.freemathhelp.com/forum/th...Before-Posting!!

 

[stapel]

What is the limit for this sequence?

\(\displaystyle (f)\, a_n\, =\, \begin{cases} e^n, &\mbox{ for }\, 0\, \leq\, n\, < 100 \\ e^{-n}, &\mbox{ for }\, n\, \geq\, 100 \end{cases}\)

the answer says its 0. But, I'm convinced the limit doesn't exist.

I'm assuming that you're taking the limit as "n" "goes to infinity"...? If so, then what value or expression do you get for the following?

. . . . .\(\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\, \)\(\displaystyle \dfrac{1}{e^n}\, \mbox{ for }\, n\, \geq\, 100\)

When you reply, please show your steps and your reasoning. Thank you!

 

[Steven G]

What is the limit for this sequence?

\(\displaystyle (f)\, a_n\, =\, \begin{cases} e^n, &\mbox{ for }\, 0\, \leq\, n\, < 100 \\ e^{-n}, &\mbox{ for }\, n\, \geq\, 100 \end{cases}\)

the answer says its 0. But, I'm convinced the limit doesn't exist.

OK, so let's start off by you telling us why the limit does not exist. I actually agree with you since you were not given which limit you are looking at (is n approaching pi or 17.3 or infinity)

 

[pka]

I actually agree with you since you were not given which limit you are looking at

@Jomo, why in the world would you ask that if you understand anything about sequences?
A sequence is a function from the positive integers.
As such it is absolutely absurd to think \(\displaystyle n\to\pi\). Only \(\displaystyle n\to\infty\) is possible
Moreover, each point of the graph of a sequence is an isolated point if it is not a limit point.
If a sequence has a unique limit point that is the limit of the sequence as \(\displaystyle n\to\infty\).
So you see the limiting value of \(\displaystyle n\) is \(\displaystyle \infty\) by definition.

 

[Steven G]

@Jomo, why in the world would you ask that if you understand anything about sequences?
A sequence is a function from the positive integers.
As such it is absolutely absurd to think \(\displaystyle n\to\pi\). Only \(\displaystyle n\to\infty\) is possible
Moreover, each point of the graph of a sequence is an isolated point if it is not a limit point.
If a sequence has a unique limit point that is the limit of the sequence as \(\displaystyle n\to\infty\).
So you see the limiting value of \(\displaystyle n\) is \(\displaystyle \infty\) by definition.

Yeah, I know that the limit can't be as n goes to pi. I was thinking that the op did not know that n was going to infinity as I can't imagine that they would not then see that the limit is 0

 

[wolf441]

OMG I just realized how dumb the question I asked was. Sorry guys I wasnt looking at it in a piece wise way. I was looking at two different sequences. Thanks for your help guys!!