Solve the limit problem

Feb 2018
63
3
Iran
20200215_175842.jpg
How am I supposed to solve this problem??
 

romsek

Math Team
Sep 2015
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1,673
USA
As written, it makes no sense.

If the limit were as $n \to \infty$ then divide top and bottom by $n^2$ to obtain

\(\displaystyle \frac{\text{a bunch of stuff over $n^2$ that goes to zero} + 2n^2}{\text{a few things over $n^2$ that go to zero} + 7n^2} = \frac 2 7\)
 

SDK

Sep 2016
797
541
USA
As written, it makes no sense.

If the limit were as $n \to \infty$ then divide top and bottom by $n^2$ to obtain

\(\displaystyle \frac{\text{a bunch of stuff over $n^2$ that goes to zero} + 2n^2}{\text{a few things over $n^2$ that go to zero} + 7n^2} = \frac 2 7\)
This doesn't work since the number of things you are neglecting in the numerator is also growing with $n$. It's not hard to prove that the numerator is proportional to a cubic polynomial in $n$ (it even has a closed form), so the limit would be $\infty$.
 
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romsek

Math Team
Sep 2015
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USA
This doesn't work since the number of things you are neglecting in the numerator is also growing with $n$. It's not hard to prove that the numerator is proportional to a cubic polynomial in $n$ (it even has a closed form), so the limit would be $\infty$.
oops.

Yes, numerator is \(\displaystyle \frac 1 6(n + 3n^2 + 2n^3)\)
 

skipjack

Forum Staff
Dec 2006
21,478
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No, the numerator is ${\large\frac16}(2n^3 + 9n^2 + 13n + 6)$.
 

romsek

Math Team
Sep 2015
2,958
1,673
USA
No, the numerator is ${\large\frac16}(2n^3 + 9n^2 + 13n + 6)$.
if you stop at (n+1) yes, which I suppose I should have done to avoid confusion, but since the limit is in n, I just summed to n.

It's all the same since you let n go to infinity.