Calculating a limit of a sequence

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Hi guys,

Homework Statement


I'm trying to calculate the limit of the following sequence:
an:= (\frac{n^{2}+10}{n^{2}-5n})^{(3n^{2}+2)}​

2. The attempt at a solution
Ok so I got this so far:

an = \left(\frac{\left(n^2\left(1+\frac{10}{n^2}\right)\right)}{\left(n^2\left(1-\frac{5}{n}\right)\right)}\right)^{n^{2}(3+\frac{2}{n})}​

\lim_{n\to \infty } an = \lim_{n\to \infty } 1^{∞}​

So this is what I got till now and I don't know really how to proceed or how can I approach this problem differently to find the correct solution which is ∞.

Any help, tips are welcome.
 
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lazy.man said:
Hi guys,

Homework Statement


I'm trying to calculate the limit of the following sequence:
an:= (\frac{n^{2}+10}{n^{2}-5n})^{(3n^{2}+2)}​

2. The attempt at a solution
Ok so I got this so far:

an = \left(\frac{\left(n^2\left(1+\frac{10}{n^2}\right)\right)}{\left(n^2\left(1-\frac{5}{n}\right)\right)}\right)^{n^{2}(3+\frac{2}{n})}​

\lim_{n\to \infty } an = \lim_{n\to \infty } 1^{∞}​

So this is what I got till now and I don't know really how to proceed or how can I approach this problem differently to find the correct solution which is ∞.

Any help, tips are welcome.
Without having worked the problem out, my first thought is to find the limit of the log of this.

If \displaystyle \lim_{n\to\infty}\ln(a_n)=L\,,\ \text{ then } \lim_{n\to\infty}a_n=e^L
 
I don't know if this is allowed with products of limits rule but:

3n^2 \ln \left( \frac{n^2}{n-5n}\right) = 3n \ln \left[\left( 1 +\frac{5}{n} \right)^n \right] = 3n \cdot 5 = 15n \rightarrow \infty
 
lazy.man said:
Hi guys,

Homework Statement


I'm trying to calculate the limit of the following sequence:
an:= (\frac{n^{2}+10}{n^{2}-5n})^{(3n^{2}+2)}​

2. The attempt at a solution
Ok so I got this so far:

an = \left(\frac{\left(n^2\left(1+\frac{10}{n^2}\right)\right)}{\left(n^2\left(1-\frac{5}{n}\right)\right)}\right)^{n^{2}(3+\frac{2}{n})}​

\lim_{n\to \infty } an = \lim_{n\to \infty } 1^{∞}​

So this is what I got till now and I don't know really how to proceed or how can I approach this problem differently to find the correct solution which is ∞.

Any help, tips are welcome.

Binomial theorem.

{(\frac{n^{2}+10}{n^{2}-5n})}^{(3n^{2}+2)} = {(1 + \frac{5}{n} + \frac{35}{n(n-5)})}^{(3n^2 + 2)} > {(1 + \frac{5}{n})^{3n^2} = {[{(1+\frac{5}{n})}^n}]}^{3n}, the limit of which is e^{15n}, which is ∞.

This is a slight (more direct) variation of dirk_mec's idea.

EDIT: BTW, I messed up the sign, clearly it should be greater than ('>').
 
Last edited:
Thank you guys for your help and I'm posting my solution to the problem.

newline\newline\newline&space;\lim_{n&space;\to&space;\infty}&space;e^{15n}&space;=&space;\infty.gif
 
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