Re: Sequences, convergence

Is the limit:

$\displaystyle \lim_{x\to \infty} \frac{3n^2+4^n+2^n}{(n^{23})^n+4n^7}$

?

Re: Sequences, convergence

Sorry the it should read n^2*3^n not (n^23)^n.

Re: Sequences, convergence

Re: Sequences, convergence

$\displaystyle \displaystyle \lim_{n\to\infty} \frac{3n^2+4^n+2^n}{n^2 \, 3^n +4n^7}$

Start by dividing top & bottom by $\displaystyle 4^n$

Re: Sequences, convergence

I did this and got 3n^2/4^n + 1 + (2/4)^n. I am not sure what to do next.

Re: Sequences, convergence

You should get : $\displaystyle \displaystyle \lim_{n\to\infty} \frac{ \frac{3n^2}{4^n} + 1 + \left( \frac{1}{2} \right)^n } { n^2 \left( \frac{3}{4} \right)^n + \frac{4n^7}{4^n} }$

You need two things here :

(1) L'Hospital's Rule.

(2) $\displaystyle \lim_{n\to\infty} r^n = 0$ if $\displaystyle |r| < 1$.