How can $\lim_{n\to \infty} (3^n + 4^n)^{1/n} = 4$? If $\lim_{n\to \infty} (3^n + 4^n)^{1/n} = 4$, then $\lim_{n\to \infty} 3^n + 4^n=\lim_{n\to \infty}4^n$ which implies that $\lim_{n\to \infty} 3^n=0$ which is clearly not correct. I tried to do the limit myself, but I got $3$. The way I did is that at the step $\lim_{n\to \infty} 3^n + 4^n=\lim_{n\to \infty}L^n$ I divided everything by $4^n$, and got $\lim_{n\to \infty} (\frac{3}{4})^n + 1=\lim_{n\to \infty} (\frac{L}{4})^n$. Informally speaking, the $1$ on the LHS is going to be very insignificant as $n \rightarrow \infty$, so $L$ would have to be $3$. Could someone explain to me why am I wrong and how can the limit possibly be equal to $4$? Thanks!

$\infty-\infty$ is not well-defined.
