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!
$\displaystyle \lim_{n \to \infty} \left(2^n + 3^n\right)^{\frac2{n}}$
$\displaystyle =\lim_{n \to \infty} \left(3^n\left(\frac{2^n}{3^n} + 1\right)\right)^{\frac2{n}}$
$\displaystyle =\lim_{n \to \infty} 3^2 \left(\frac{2^n}{3^n} + 1\right)^{\frac2{n}}$
$\displaystyle =\lim_{n \to \infty} 3^2 \left(\left(\frac{2}{3}\right)^n + 1\right)^{\frac2{n}}$
$\displaystyle = 9 \,\,\,\,\,\left[\because n \to \infty \Rightarrow \left(\frac{2}{3}\right)^n \to 0\right]$
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}} $
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