# 2 limit questions

by cpyap
Tags: limit
 P: 13 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!
 Mentor P: 4,177 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
P: 13
 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.

P: 13

## 2 limit questions

 P: 867 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: $$\lim_{x\rightarrow \infty}\ln\left[\left(1 + \frac{1}{x}\right)^x(x + 1)\right] - 1 = \lim_{x\rightarrow \infty}\ln\left(1 + \frac{1}{x}\right)^x + \lim_{x\rightarrow \infty}\ln(x + 1) - 1 = \ln\left(\lim_{x\rightarrow \infty}\left(1 + \frac{1}{x}\right)^x\right) - 1 + \lim_{x\rightarrow \infty}\ln(x + 1)$$
 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: $$\lim_{x\rightarrow \infty}\ln\left[\left(1 + \frac{1}{x}\right)^x(x + 1)\right] - 1 = \lim_{x\rightarrow \infty}\ln\left(1 + \frac{1}{x}\right)^x + \lim_{x\rightarrow \infty}\ln(x + 1) - 1 = \ln\left(\lim_{x\rightarrow \infty}\left(1 + \frac{1}{x}\right)^x\right) - 1 + \lim_{x\rightarrow \infty}\ln(x + 1)$$