Limit of n/(n+1)^3 as n Approaches Infinity | Quick Homework Question

[michonamona]

Homework Statement

limit of n/(n+1)^3 as n approaches infinity

Homework Equations

The Attempt at a Solution

The degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator. Is it correct to conclude that the limit of this sequence, as n approaches infinity is zero?

I appreciate your help

 

[Landau]

Yes.

 

[Mark44]

Homework Statement

limit of n/(n+1)^3 as n approaches infinity

Homework Equations

The Attempt at a Solution

The degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator. Is it correct to conclude that the limit of this sequence, as n approaches infinity is zero?

I appreciate your help

Yes.
You can do this by factoring n^3 out of the denominator, to get:
\lim_{n \to \infty}\frac{n}{(n + 1)^3} = \lim_{n \to \infty}\frac{n}{n^3(1 + 1/n)^3}

 

[Ksitov]

Hi !

When you have a problem with a limit.

It's necessary to think of factorizing, it's the method !

You factorise numerator, denominator and you simplify !

Sorry for my english level, I'm new and french !

 

[Gib Z]

Hi !

When you have a problem with a limit.

It's necessary to think of factorizing, it's the method !

You factorise numerator, denominator and you simplify !

That's a good method when you have the ratio of two polynomials and they are of the form 0/0. eg If you had \lim_{x\to 2}\frac{x^2-4}{x-2}, then when you put in x=2, it is 0/0, so to get that limit, you could use your method. But with limits as x goes to infinity, you should dividing top and bottom by the highest power of x, as Mark44 did.

Sorry for my english level, I'm new and french !

Welcome to PF, and your English is fine :)

 

[Ksitov]

Another technique, you can verify with calculator the result.

But it's necessary to demonstrate with the calcul !

Good Bye

 

[System]

Another one:
put t=n+1
Clearly, t goes to infinity as n goes to infinity,
so the limit will be :
lim of (t-1)/t^3 as t goes to infinity
Devide top and bottom by t:
lim of (1 - (1/t) )/t^2 as t goes to infinity = (1-0)/infinity=1/infinity=0

 

[njama]

Another one:

Use L'Hopital rule.

 

[Landau]

Another technique, you can verify with calculator the result.

But it's necessary to demonstrate with the calcul !

Of course a calculator can't verify this, only make it plausible, but I don't think you meant that.

 

[statdad]

One more approach:

<br /> 0 &lt; \frac n {(n+1)^3} &lt; \frac n {n^3} = \frac 1 {n^2}<br />

and your sequence is trapped between 0 and another sequence that converges to zero.