2 limit questions

cpyap

1. The problem statement, all variables and given/known data
What will be the limit of
a) [tex]lim_{n\rightarrow\infty}\frac{2+3^{n+1}}{(n+1)(2+3^{n})}[/tex]
b) [tex]lim_{x\rightarrow\infty} (x+1)ln(x+1) - x ln(x) - 1[/tex]



2. Relevant equations
-

3. The attempt at a solution
It is actually coming from infinite series question,
which I goes into the limit step and stuck.
[tex]lim_{n\rightarrow\infty}\frac{2+3^{n+1}}{(n+1)(2+3^{n})}[/tex] look like equal to 0,
[tex]ln\left[\frac{(x+1)^{x+1}}{x^{x}}\right][/tex] look like equal to [tex]\infty[/tex],
but I don't know how to work on them,
which pretty much stuck on this stage.
Any help will be appreciated.
Thanks!

Office_Shredder

For the first one, try dividing the numerator and denominator by [tex]3^n[/tex], and see if you can get an idea for what the numerator and denominator look like when n is large

cpyap

[tex]lim_{n\rightarrow\infty}\frac{\frac{2}{3^{n}}+3}{\frac{2n+2}{3^{n}}+n+1 } = \frac{3}{\infty+1} = 0
[/tex]
I believe I'm getting tired,
missed a whole point here.
Thanks a lot for help.

cpyap

How about b)?

cpyap

Question edited. Just try to make the question b) clearer.
Thanks!

Bohrok

You can rewrite b) to make taking the limit easier:

[tex]\frac{(x+1)^{x+1}}{x^{x}} = \frac{(x + 1)^x(x+1)}{x^x} = \left(\frac{x+1}{x}\right)^x(x + 1) = \left(1 + \frac{1}{x}\right)^x(x + 1)[/tex]

Then taking the limit of the expression:

[tex]\lim_{x\rightarrow \infty}\ln\left[\left(1 + \frac{1}{x}\right)^x(x + 1)\right] - 1 = \lim_{x\rightarrow \infty}\ln\left(1 + \frac{1}{x}\right)^x + \lim_{x\rightarrow \infty}\ln(x + 1) - 1 = \ln\left(\lim_{x\rightarrow \infty}\left(1 + \frac{1}{x}\right)^x\right) - 1 + \lim_{x\rightarrow \infty}\ln(x + 1)[/tex]

Dick

You don't even have to work that hard. Since x->infinity you can take x>1. So (x+1)/x>=1 and (x+1)^x/x^x>=1. The leftover (x+1) goes to infinity. It's basically a comparison test.

Bohrok

You're very right, I know it's more than one might do for this problem. But since there was a -1 in the expression, and I saw it as ln(lim_x→∞(1 + 1/x)^x), that limit gets rid of the -1 and leaves you with just a nice lim_x→∞ ln(x + 1).