Can we solve this limit without squeeze theorem I was wondering if we can solve this limit without using squeeze (sandwich) theorem.
$$\lim_{n\to \infty}(3^n+5^n)^{2/n}$$
 A: Certainly.
$$
\lim_{n \to +\infty} \exp\left(\ln\left((3^n + 5^n)^{2/n}\right)\right) = \lim_{n \to +\infty} \exp \left(\frac{2\ln(3^n + 5^n)}{n}\right) = \exp\left(\lim_{n \to +\infty}\frac{2\ln(3^n + 5^n)}{n}\right)
$$
Now, let's solve $\displaystyle \lim_{n \to +\infty}\frac{2\ln(3^n + 5^n)}{n}$. We have $\displaystyle \frac{+\infty}{+\infty}$, so we can apply l'Hospital's rule straight away, giving us:
$$
\lim_{n \to +\infty}\frac{2 \ln(3) 3^n + 2 \ln(5) 5^n}{3^n + 5^n} = 2 \ln(5)
$$
So, the limit is $\displaystyle \exp(2 \ln(5)) = 25$.
A: $$\lim_{n \to \infty} f(x) =  e^{\lim_{n \to \infty}{\ln(f(x))}}=e^{\lim_{n \to \infty}{\frac2n\ln(3^n+5^n)}}=e^{2\lim_{n \to \infty}{\frac{\ln{3^n+5^n}}{n}}}$$
Now let's use a L'Hopital's rule:
$$e^{2\lim_{n \to \infty}{\frac{\ln{3^n+5^n}}{n}}}=e^{2\lim_{n \to \infty}{\frac{3^n\ln3+5^n\ln5}{3^n+5^n}}}=e^{2\lim_{n \to \infty}{\ln5+\frac{3^n\ln3-3^n\ln5}{3^n+5^n}}}=e^{2\ln5}=e^{\ln25}=25$$
A: $$\begin{align}
&\lim_{n\to \infty}(3^n+5^n)^{2/n}\\
=&\lim_{n\to \infty}\exp\left( {2\log(3^n+5^n)/n}\right)\\
=&\exp\left({\lim_{n\to \infty}{2\log(3^n+5^n)/n}}\right)\\
=&\exp\left({\lim_{n\to \infty}{\frac{2\log(3)3^n+2\log(5)5^n}{3^n+5^n}}}\right)\\
=&\exp\left({\lim_{n\to \infty}{\frac{2\log(3)(3/5)^n+2\log(5)}{(3/5)^n+1}}}\right)\\
=&\exp\left({{\frac{[0]+2\log(5)}{[0]+1}}}\right)\\
=&\exp\left(2\log 5\right)\\
=&25\\
\end{align}$$
A: You may just write, as $n$ tends to $+\infty$,
$$
\begin{align}
(3^n+5^n)^{2/n}&=e^{\frac2n\log \left(3^n+5^n \right)}\\\\
&=e^{\frac2n\log \left(5^n\right)+ \frac2n\log \left(1+(3/5)^n \right)}\\\\
&=e^{2\log 5+ \frac2n \log \left(1+(3/5)^n \right)}\\\\
&=e^{2\log 5+ \frac2n (3/5)^n }\\\\
&\sim e^{2\log 5}\times e^0\\\\
&\sim25
\end{align}
$$ where we have used 
$$
\log (1+x)\sim_0 x
$$ with $x:=(3/5)^n$, $n$ being great.
A: $$\lim_{n \to \infty }\sqrt[n]{(3^n+5^n)^2}=\\\lim_{n \to \infty }\sqrt[n]{(5^n((\frac{3}{5})^n+1))^2}=\\5^2\lim_{n \to \infty }\sqrt[n]{((\frac{3}{5})^n+1)^2}=\\$$as we now $$\lim_{n \to \infty }(\frac{3}{5})^n=0$$so $$5^2\lim_{n \to \infty }\sqrt[n]{((\frac{3}{5})^n+1)^2}=5^2*\sqrt[n]{(0+1)^2}=25$$
