Limits of sequences as x heads to infinity

[BigJon]

cn= (4n)/(n+4n^(1/n))

When i set it up i think i should use l'hopital but I am confused what to do with the 4n^(1/n) term.

an=(7^(2n))/(n!)

I know this is a geometric sequence and top and bottom increase initially then tend to 0, but I am lost on how to show the work. should i expand the factorial or how would i use this with the squeeze theorem

 

[Ray Vickson]

cn= (4n)/(n+4n^(1/n))

When i set it up i think i should use l'hopital but I am confused what to do with the 4n^(1/n) term.

an=(7^(2n))/(n!)

I know this is a geometric sequence and top and bottom increase initially then tend to 0, but I am lost on how to show the work. should i expand the factorial or how would i use this with the squeeze theorem

Does 4n^(1/n) mean 4[n^(1/n)], or does it mean (4n)^(1/n). If one reads it using standard rules, what you have written is 4[n^(1/n)]. So, what is the limit of n^(1/n)? [Hint: take logarithms.]

RGV

 

[BigJon]

This is what it looks like sorry for confusion

 

[Ray Vickson]

This is what it looks like sorry for confusion

There was no actual confusion; what you wrote is what you meant. (However, that is often not the case on this Forum, so that is why I asked.)

Anyway, what is your answer to my question about n^(1/n) as n → ∞?

RGV

 

[BigJon]

ln(n^(1/n))
(1/n)ln(n)=ln(n)/n

lim x->infin of ln(n)/n ->0 using l'hopitals rule

I don't understand did you apply ln to all the terms?

 

[Ray Vickson]

ln(n^(1/n))
(1/n)ln(n)=ln(n)/n

lim x->infin of ln(n)/n ->0 using l'hopitals rule

I don't understand did you apply ln to all the terms?

I did not apply ln to anything---you did. So, given what you just did above, what can you say about whether or not n^(1/n) has a limit, and if so, what that limit is? Why does the answer to this question simplify the original problem?

RGV

 

[BigJon]

So i put it into the form of lnx/x use lhopitals then it goes to 4/1+((1/x)/(x)) limit of this as it goes to inifintiny becomes 4

 

[Ray Vickson]

So i put it into the form of lnx/x use lhopitals then it goes to 4/1+((1/x)/(x)) limit of this as it goes to inifintiny becomes 4

Once you have the form above, you do not need l'Hospital any more. Do you see why?

RGV

 

[BigJon]

Well i already took it once so i don't need to take it again. i just take limits after that.


What i need to know is how can you apply the Ln to just n^(1/n) without applying it to the other n in the denominator or 4n in the numerator.

 

[Ray Vickson]

Well i already took it once so i don't need to take it again. i just take limits after that.


What i need to know is how can you apply the Ln to just n^(1/n) without applying it to the other n in the denominator or 4n in the numerator.

Your denominator has two terms: n and n^(1/n). You need to know how the denominator behaves for large n, and that has nothing at all to do with l'Hospital's rule or anything like it. How does n behave when n → ∞? How does n^(1/n) behave when n → ∞? Do, how does the denominator behave when n → ∞? Furthermore, if you re-write your fraction as
$$\frac{4}{1 + r(n)},$$
where
$$r(n) = \frac{1}{n} n^{1/n},$$
it is easy to see what is the limit of r(n) as n → ∞. If that limit is L, the whole fraction has limit 4/(1+L).

Just as you do not need l'Hospital to do lim (2x)/x [because for x ≠ 0 it is just equal to 2], so you do not need l'Hospital here.

RGV