1. ## limits

Hi i need help finding limts!
(2(^n)+ 3 (^n))^2/n
from this i am suppose to get
3^2(1 +2/3^n)^(2/n)
Do not know how to get there though. The eventual limit is 9!

2. Originally Posted by studentsteve1202
Hi i need help finding limts!
(2(^n)+ 3 (^n))^2/n
from this i am suppose to get
3^2(1 +2/3^n)^(2/n)
Do not know how to get there though. The eventual limit is 9!
$\displaystyle (2^n+3^n)^{2/n}$

$\displaystyle [2^n(1+(3/2)^n)]^{2/n}$

because $\displaystyle 3^n = 2^n * (3/2)^n$

3. Originally Posted by studentsteve1202
Hi i need help finding limts!
(2(^n)+ 3 (^n))^2/n
from this i am suppose to get
3^2(1 +2/3^n)^(2/n)
Do not know how to get there though. The eventual limit is 9!
$\displaystyle \lim_{n \to \infty} \left(2^n + 3^n\right)^{\frac2{n}}$

$\displaystyle =\lim_{n \to \infty} \left(3^n\left(\frac{2^n}{3^n} + 1\right)\right)^{\frac2{n}}$

$\displaystyle =\lim_{n \to \infty} 3^2 \left(\frac{2^n}{3^n} + 1\right)^{\frac2{n}}$

$\displaystyle =\lim_{n \to \infty} 3^2 \left(\left(\frac{2}{3}\right)^n + 1\right)^{\frac2{n}}$

$\displaystyle = 9 \,\,\,\,\,\left[\because n \to \infty \Rightarrow \left(\frac{2}{3}\right)^n \to 0\right]$

4. $\displaystyle \left( {2^n + 3^n } \right)^{2/n} = \left( {\sqrt[n]{{2^n + 3^n }}} \right)^2 \to 9$

$\displaystyle 0 < a < b\quad \Rightarrow \quad \left( {\sqrt[n]{{a^n + b^n }}} \right) \to b$

5. Hello, studentsteve1202!

$\displaystyle \left(2^n+ 3^n\right)^{\frac{2}{n}}$

from this i am suppose to get: .$\displaystyle 3^2\cdot\left[1 + \left(\frac{2}{3}\right)^n\right]^{\frac{2}{n}}$

We have: .$\displaystyle \left(2^n + 3^n\right)^{\frac{2}{n}}$

Multiply the first term by $\displaystyle \frac{3^n}{3^n}\!:\quad\left({\color{blue}\frac{3^ n}{3^n}}\cdot2^n + 3^n\right)^{\frac{2}{n}} \;=\;\left(3^n\cdot\frac{2^n}{3^n} + 3^n\right)^{\frac{2}{n}}$

Factor: . $\displaystyle \left[3^n\left(\frac{2^n}{3^n} + 1\right)\right]^{\frac{2}{n}} \;=\;\left(3^n\right)^{\frac{2}{n}}\left[\left(\frac{2}{3}\right)^n + 1\right]^{\frac{2}{n}} \;=\;3^2\cdot\left[\left(\frac{2}{3}\right)^n + 1\right]^{\frac{2}{n}}$

6. Given two positive real numbers $\displaystyle a$ and $\displaystyle b$

We have$\displaystyle \lim_{n \to +\infty}(a^n+b^n)^{\frac{1}{n}}=\max(a;b)$ (1)

The proof of this is almost the same Isomorphism provided.

And:$\displaystyle \lim_{x \to -\infty}(a^x+b^x)^{\frac{1}{x}}=\min(a;b)$ (you can show this limit by applying (1) )

Interesting

7. Originally Posted by studentsteve1202
Hi i need help finding limts!
(2(^n)+ 3 (^n))^2/n
from this i am suppose to get
3^2(1 +2/3^n)^(2/n)
Do not know how to get there though. The eventual limit is 9!
A more inelegant way of doing this since everyone else did it the fun way would be to take the ln of both sides and then use L'hopitals

8. Thankyou very much everyone. Very helpful