1. ## Limits

I have difficulties in solving problems

1) Show that lim (1 + (1/2n))^n = sqrt e.
n-> oo

2)find lim (1+2/n)^n
n-> oo

3) lim (1 + 3/n) ^n
n-> oo

4) suppose 0 < Sn and Sn+1 < rSn, where r is a constant such that 0 < r < 1. Show that limn->oo Sn = 0.

5) suppose 0 < Sn and Sn+1 >= rSn, where r is a constant such that r > 1. Show that limn->oo Sn = +oo.

2. Originally Posted by mathsohard
I have difficulties in solving problems

1) Show that lim (1 + (1/2n))^n = sqrt e.
n-> oo
$\displaystyle \displaystyle\lim_{n \to \infty}\left[1+\frac{1}{2n}\right]^n=\lim_{n \to \infty}\left(\left[1+\frac{1}{2n}\right]^{2n}\right)^{\frac{1}{2}}$

which is ?

3. For #4, show that $\displaystyle S_{n+1}<r^nS_1$

4. Can you give me some help on showing that???

5. Originally Posted by mathsohard
I have difficulties in solving problems

1) Show that lim (1 + (1/2n))^n = sqrt e.
n-> oo

2)find lim (1+2/n)^n
n-> oo

3) lim (1 + 3/n) ^n
n-> oo

4) suppose 0 < Sn and Sn+1 < rSn, where r is a constant such that 0 < r < 1. Show that limn->oo Sn = 0.

5) suppose 0 < Sn and Sn+1 >= rSn, where r is a constant such that r > 1. Show that limn->oo Sn = +oo.
Pretty much for 1, 2, 3:
$\displaystyle \displaystyle $\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{C}{n}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {e^{n \cdot \ln \left( {1 + \frac{C}{n}} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } n \cdot \ln \left( {1 + \frac{C}{n}} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } n \cdot \frac{C}{n}}} = {e^C}$$

6. for 2 and 3 if I wanted to use the fact (1+2/n) = (1+ 1/(n+1))(1+ 1/n) how can I do this??

7. I'd have approached (2) differently:

$\displaystyle \displaystyle\lim_{n \to \infty}\left[1+\frac{2}{n}\right]^n=\lim_{n \to \infty}\left(\left[1+\frac{1}{\left(\frac{n}{2}\right)}\right]^{\frac{n}{2}\right)^2$

8. Thanks for 1, 2, 3 , I still need help on 4, 5

9. Originally Posted by mathsohard
Thanks for 1, 2, 3 , I still need help on 4, 5
Referring to Plato's post:

$\displaystyle S_1>0,\;\;0<r<1$

$\displaystyle S_2<rS_1\Rightarrow\ S_3<rS_2\Rightarrow\ S_3<r^2S_1$

$\displaystyle S_4<r^3S_1$

$\displaystyle S_{n+1}<r^nS_1$

Since $\displaystyle 0<r<1,\;\;\;\displaystyle\lim_{n \to \infty}r^n=\;?$

10. Thank you very much

11. $\displaystyle 0<r<1\Rightarrow\ r=\frac{a}{b}=\frac{a}{ka}$

since $\displaystyle b>a\Rightarrow\ b=ka,\;\;k>1$

$\displaystyle \displaystyle\Rightarrow\lim_{n \to \infty}\left(\frac{1}{k}\right)^n=0,\;\;k>1$

Or:

$\displaystyle \displaystyle\ 0<r<1\Rightarrow\ k=\frac{1}{r}\Rightarrow\ k>1$

$\displaystyle \displaystyle\ r^n=\left(\frac{1}{k}\right)^n$ etc