
#1
Sep210, 02:30 PM

P: 13

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! 



#2
Sep210, 02:36 PM

Mentor
P: 4,499

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




#3
Sep210, 02:43 PM

P: 13

[/tex] I believe I'm getting tired, missed a whole point here. Thanks a lot for help. 



#4
Sep210, 03:15 PM

P: 13

2 limit questions
How about b)?




#5
Sep210, 04:11 PM

P: 13

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



#6
Sep210, 09:49 PM

P: 867

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] 



#7
Sep210, 10:03 PM

Sci Advisor
HW Helper
Thanks
P: 25,168





#8
Sep310, 12:01 AM

P: 867

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).



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