I don't want to use the definition, is there an easier way?
Like, in the case of limits of functions, it is pretty easy to say by the composition theorem.
Is there any such theorem in case of sequences too?
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A sequence is a function. – Arnaud Mortier Aug 17 '18 at 05:35
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@ArnaudMortier, sequence is a function from $\mathbb{N}$. But the theorem is for continuous functions only. So how can you talk about continuity when your domain is $\mathbb{N}$? – Aditya Agarwal Aug 17 '18 at 05:38
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Now that would make a good question, although it is probably a duplicate. – Arnaud Mortier Aug 17 '18 at 05:40
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@ArnaudMortier, if its duplicate. Please mark it as duplicate. But my answer answers my question. – Aditya Agarwal Aug 17 '18 at 05:41
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Here: https://math.stackexchange.com/questions/2881876/can-a-function-with-infinitely-many-holes-in-the-domain-still-be-continuous/2881901#2881901 – Arnaud Mortier Aug 17 '18 at 05:48
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Hint:
As exponential function is continuous and "limit of the function"=function of the limit" for a continuous function, so$$\lim_n e^{1/n}=e^{\lim_n 1/n}=e^0=1.$$
Empty
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Your answer is not at all legible. I am asking about sequences. So I presume, that limit, in this case means the number to which the sequence is convergent. But then how could you take the limit in the exponent? – Aditya Agarwal Aug 17 '18 at 05:21
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This is not an acceptable answer. More effort is needed in terms of formatting and clearly explaining the concept in words – John Zimmerman Aug 17 '18 at 05:21
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Thats what I am asking. My question is about sequences, not functions. – Aditya Agarwal Aug 17 '18 at 05:22
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@Empty, I can't still see the hint. It clearly restates what I have not asked you to state, that is limit of functions. I am talking about limit of sequences. – Aditya Agarwal Aug 17 '18 at 05:25
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@AdityaAgarwal Do you know, $\displaystyle \lim_{x\to a} f(g(x))=f(\lim_{x\to a}g(x))$, if $f$ is a continuous function ?? – Empty Aug 17 '18 at 05:29
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Yup, but not for a sequence. See my answer. Thanks for answering, but it was just a case of misunderstanding. – Aditya Agarwal Aug 17 '18 at 05:31
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@AdityaAgarwal What's your problem? Take $f(x)=e^x$ and $g(x)=1/x$ and take the limit as $x\to \infty$ and use my 2nd last comment to evaluate the limit – Empty Aug 17 '18 at 05:34
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If you still unable to understand then please wait for another answer from other members..!! – Empty Aug 17 '18 at 05:37
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@Empty. sequence is a function from N. But the theorem is for continuous functions only. So how can you talk about continuity when your domain is N? – Aditya Agarwal Aug 17 '18 at 05:39
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I've already stated in my above comment , which you have to take $f$ and $g$. Do you think $f$ is not continuous on $\Bbb R$ ? – Empty Aug 17 '18 at 05:40
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I cannot continue this discussion here, I am only going to reply on the chat now. – Aditya Agarwal Aug 17 '18 at 05:42
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I figured it out. Since the function whose domain is $\mathbb{R}$ converges to $1$ (By composiotion theorem). The sequence of the same functional rule should, too.
Since, we know that for every $\epsilon>0$, there exists $N$, so that $|f(n)-L|<\epsilon$ for $n>N$, So for every $\epsilon>0$, there exists $N$, so that $|f(n)-L|<\epsilon$ for $n \in \mathbb{I}>N$. So the sequence converges too. That is, $$\lim_{n\to\infty}a_n=L$$ too.
Aditya Agarwal
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If anything is wrong with my answer, or is irrelevant, please explain, the downvoter. – Aditya Agarwal Aug 17 '18 at 05:46
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For $n\in \Bbb Z^+$ let $e^{1/n}=1+A_n.$ We have $A_n>0.$ From the Binomial Theorem we have $$e=(1+A_n)^n=\sum_{j=0}^n\binom {n}{j}(A_n)^j\geq \sum_{j=0}^1\binom {n}{j}(A_n)^j=1+nA_n.$$ Therefore $\frac {e-1}{n}\geq A_n.$ This also works if we replace $e$ with any number greater than $1.$
DanielWainfleet
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