## 2 limit questions

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!

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 Blog Entries: 1 Recognitions: Homework Help 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

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

## 2 limit questions

 Question edited. Just try to make the question b) clearer. Thanks!
 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: $$\lim_{x\rightarrow \infty}\ln\left[\left(1 + \frac{1}{x}\right)^x(x + 1)\right] - 1 = \lim_{x\rightarrow \infty}\ln\left(1 + \frac{1}{x}\right)^x + \lim_{x\rightarrow \infty}\ln(x + 1) - 1 = \ln\left(\lim_{x\rightarrow \infty}\left(1 + \frac{1}{x}\right)^x\right) - 1 + \lim_{x\rightarrow \infty}\ln(x + 1)$$

Recognitions:
Homework Help
 Quote by 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: $$\lim_{x\rightarrow \infty}\ln\left[\left(1 + \frac{1}{x}\right)^x(x + 1)\right] - 1 = \lim_{x\rightarrow \infty}\ln\left(1 + \frac{1}{x}\right)^x + \lim_{x\rightarrow \infty}\ln(x + 1) - 1 = \ln\left(\lim_{x\rightarrow \infty}\left(1 + \frac{1}{x}\right)^x\right) - 1 + \lim_{x\rightarrow \infty}\ln(x + 1)$$