40. $\displaystyle \lim_{x\to\infty}(2^n + 3^n)^\frac{1}{n}$
The books says to try the sandwich theorem, but I have no idea what I am doing anymore. Been doing homework for like 48 hours straight .
Thanks for help
$\displaystyle \displaystyle \lim_{n \to \infty}(2^n + 3^n)^{\frac{1}{n}} = \lim_{n \to \infty}e^{\ln{(2^n + 3^n)^{\frac{1}{n}}}}$
$\displaystyle \displaystyle = e^{\lim_{n \to \infty}\ln{(2^n + 3^n)^{\frac{1}{n}}}}$
$\displaystyle \displaystyle = e^{\lim_{n \to \infty}\frac{1}{n}\ln{(2^n + 3^n)}}$
$\displaystyle \displaystyle = e^{\lim_{n \to \infty}\frac{\ln{(2^n + 3^n)}}{n}}$
$\displaystyle \displaystyle = e^{\lim_{n \to \infty}\frac{\frac{2^n\ln{2} + 3^n\ln{3}}{2^n + 3^n}}{1}}$ by L'Hospital's Rule
$\displaystyle \displaystyle = e^{\lim_{n \to \infty}\frac{2^n\ln{2} + 3^n\ln{3}}{2^n + 3^n}}$
$\displaystyle \displaystyle = e^{\lim_{n \to \infty}\frac{\frac{2^n\ln{2}}{3^n} + \ln{3}}{\frac{2^n}{3^n} + 1}}$
$\displaystyle \displaystyle = e^{\lim_{n \to \infty}\frac{\left(\frac{2}{3}\right)^n \ln{2} + \ln{3}}{\left(\frac{2}{3}\right)^n + 1}}$
$\displaystyle \displaystyle = e^{\frac{0 + \ln{3}}{0 + 1}}$
$\displaystyle \displaystyle = e^{\ln{3}}$
$\displaystyle \displaystyle = 3$.
With sandwich rule it is much simple!
Notice that:
$\displaystyle ({3^n})^\frac{1}{n}<(2^n + 3^n)^\frac{1}{n}<(2\cdot 3^n)^\frac{1}{n}$
Now:
$\displaystyle lim_{n\to\infty} ({3^n})^\frac{1}{n}=lim_{n\to\infty} 3=3$
$\displaystyle lim_{n\to\infty} ({2\cdot 3^n})^\frac{1}{n}=lim_{n\to\infty} {2}^\frac{1}{n}\cdot lim_{n\to\infty} ({3^n})^\frac{1}{n}=1\cdot 3=3$
Hence, by the sandwich rule $\displaystyle lim_{n\to\infty} (2^n + 3^n)^\frac{1}{n}=3$