2 limit questions

1. Sep 2, 2010

cpyap

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

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.
$$lim_{n\rightarrow\infty}\frac{2+3^{n+1}}{(n+1)(2+3^{n})}$$ look like equal to 0,
$$ln\left[\frac{(x+1)^{x+1}}{x^{x}}\right]$$ look like equal to $$\infty$$,
but I don't know how to work on them,
which pretty much stuck on this stage.
Any help will be appreciated.
Thanks!

Last edited: Sep 2, 2010
2. Sep 2, 2010

Office_Shredder

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

3. Sep 2, 2010

cpyap

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

4. Sep 2, 2010

cpyap

How about b)?

5. Sep 2, 2010

cpyap

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

6. Sep 2, 2010

Bohrok

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

$$\frac{(x+1)^{x+1}}{x^{x}} = \frac{(x + 1)^x(x+1)}{x^x} = \left(\frac{x+1}{x}\right)^x(x + 1) = \left(1 + \frac{1}{x}\right)^x(x + 1)$$

Then taking the limit of the expression:

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

7. Sep 2, 2010

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.

8. Sep 3, 2010

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(limx→∞(1 + 1/x)x), that limit gets rid of the -1 and leaves you with just a nice limx→∞ ln(x + 1).

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