Hi guys can you give me any hints how to touch this limit at all, I have tried to use the known limit with Euler's, but it's a wrong way. I don't need a solution just get me on the right track to the solution. Thank you in advance![math]\lim_{n\to\infty}\left(\frac{1+(99\cdot99)^\frac{1}{n}}{2}\right)^n[/math]
a tricky limit
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Dr.Peterson
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First, please clarify what you mean by [imath]99\cdot99[/imath]. Is the dot meant to be a decimal point, or does it mean [imath]99^2[/imath]? Either way, I'm not sure why it would be written that way. What is the context? That is, if this relates to a class, what topics have you been taught? If it is from somewhere else, where?
Second, please show one attempt, which will help us see both what you mean by the expression, and why you say it doesn't work. Have you tried L'Hopital's rule? What happened?
In working on it, I might replace [imath]99\cdot99[/imath] with a named constant, making it [imath]\displaystyle\lim_{n\to\infty}\left(\frac{1+a^\frac{1}{n}}{2}\right)^n[/imath], just to save some writing.
We cannot apply L'Hospitala rule to the sequences! They are not differentiable! It's the biggest mistake that people make, every mathematician will tell you that. I try to find a way without making a function of a variable 'x' and just applying L'Hospital rule like a noob. It's really tricky limit...
P. S. yeah there is 99^2
Using Dr. Peterson's suggestion, I'm getting [imath]\sqrt{a}[/imath], so [imath]99\cdot 99[/imath] is probably intentional (not a decimal point).
The way I did it:
1) express [imath]a^{\frac{1}{n}}[/imath] as an [imath]e^0[/imath]-type infinite series. That is, [imath]a^{\frac{1}{n}}=e^{\frac{1}{n}\cdot\ln a}[/imath] which has an expansion.
2) If [imath]a^{\frac{1}{n}}[/imath] is effectively [imath]1+\frac{1}{n}\cdot \ln a+\ldots[/imath], then [imath]\left(\frac{1+a^{\frac{1}{n}}}{2}\right)[/imath] becomes a [imath]\left(1+\text{something approaching zero}\right)[/imath]
3) This turns the [imath]\left(\frac{1+a^{\frac{1}{n}}}{2}\right)^n[/imath] into a [imath]1^\infty[/imath] kind of limit.
The way I did it:
1) express [imath]a^{\frac{1}{n}}[/imath] as an [imath]e^0[/imath]-type infinite series. That is, [imath]a^{\frac{1}{n}}=e^{\frac{1}{n}\cdot\ln a}[/imath] which has an expansion.
2) If [imath]a^{\frac{1}{n}}[/imath] is effectively [imath]1+\frac{1}{n}\cdot \ln a+\ldots[/imath], then [imath]\left(\frac{1+a^{\frac{1}{n}}}{2}\right)[/imath] becomes a [imath]\left(1+\text{something approaching zero}\right)[/imath]
3) This turns the [imath]\left(\frac{1+a^{\frac{1}{n}}}{2}\right)^n[/imath] into a [imath]1^\infty[/imath] kind of limit.
Dr.Peterson
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Are you sure of that?We cannot apply L'Hospital rule to the sequences! They are not differentiable! It's the biggest mistake that people make, every mathematician will tell you that. I try to find a way without making a function of a variable 'x' and just applying L'Hospital rule like a noob. It's really tricky limit...
P. S. yeah there is 99^2
If you take n as a real number, then you can find the limit, and that implies the limit of the sequence.
See here, for a fuller explanation. See Lemma 1. Certainly not all sequence limits can be solved this way, but when [imath]a_n=f(n)[/imath] for a differentiable function f, they can. Not to do this is to unnecessarily restrict yourself.
"If you take n as a real number, then you can find the limit, and that implies the limit of the sequence."
I said:
"I try to find a way without making a function of a variable 'x' and just applying L'Hospital rule"
x such that x is a real number... and then I will be able to use L'Hospital rule
I said:
"I try to find a way without making a function of a variable 'x' and just applying L'Hospital rule"
x such that x is a real number... and then I will be able to use L'Hospital rule
