Calculating limits: limit_{n -> infinity} ((n + 1)(n + 2)(n + 3))/n^3

[Qwertyuiop[]]

Hi, how would you calculate this limit:[imath]\lim _{n\to \infty }\left(\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)}{n^3}\right)[/imath] ? I used l'hopital's rule but it was very tedious taking the derivative of the numerator, is there a way to do it quicker?

 

[blamocur]

Hint: represent the fraction as a product of three fractions.

 

[lev888]

Hi, how would you calculate this limit:[imath]\lim _{n\to \infty }\left(\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)}{n^3}\right)[/imath] ? I used l'hopital's rule but it was very tedious taking the derivative of the numerator, is there a way to do it quicker?

Multiply the numerator and express the whole thing as a sum of fractions.

 

[Qwertyuiop[]]

Hint: represent the fraction as a product of three fractions.

It simplifies to [imath]\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right)[/imath] and it's easy to calculate the limit in this form which is 1. ?

 

[BigBeachBanana]

By inspection, the numerator is a polynomial degree of 3 with a coefficient of 1.
[math]\lim_{n \to \infty} = \dfrac{n^3+...}{n^3} = 1[/math]

 

[topsquark]

It simplifies to [imath]\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right)[/imath] and it's easy to calculate the limit in this form which is 1. ?

Be a tad more careful about simplifying it this way. It's slightly better to multiply it out first, then divide each term by [imath]n^3[/imath]. Any time you take the limits inside of a multiplication you could run into trouble. But for this case, yes, it does work.

Good job!

-Dan

Addendum: The example I'm thinking of you to be careful of isn't a multiplication, where you are pretty safe, but an exponential:
[imath]\displaystyle \lim_{n \to \infty} \left ( 1 + \dfrac{x}{n} \right ) ^n[/imath]

If you aren't careful, you might drop the x/n term (because it is small compared to 1 in the limit) and say that the limit is 1. The limit is actually [imath]e^x[/imath].

 

[Qwertyuiop[]]

Be a tad more careful about simplifying it this way. It's slightly better to multiply it out first, then divide each term by [imath]n^3[/imath]. Any time you take the limits inside of a multiplication you could run into trouble. But for this case, yes, it does work.

Good job!

-Dan

Addendum: The example I'm thinking of you to be careful of isn't a multiplication, where you are pretty safe, but an exponential:
[imath]\displaystyle \lim_{n \to \infty} \left ( 1 + \dfrac{x}{n} \right ) ^n[/imath]

If you aren't careful, you might drop the x/n term (because it is small compared to 1 in the limit) and say that the limit is 1. The limit is actually [imath]e^x[/imath].

ok I didn't know about this.

 

[Steven G]

It simplifies to [imath]\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right)[/imath] and it's easy to calculate the limit in this form which is 1. ?

Your method involved more algebra than needed, imo.

\(\displaystyle \dfrac{(n+1)(n+2)(n+3)}{n^3} =( \dfrac{n+1}{n})(\dfrac{n+1}{n})(\dfrac{n+3}{n})\). Now take the limit of both sides. The rhs limit will be 1*1*1=1