Is there an easy way to calculate $\lim\limits_{k \to \infty} \frac{(k+1)^5(2^k+3^k)}{k^5(2^{k+1} + 3^{k+1})}$? Is there an easy way to calculate $$\lim_{k \to \infty} \frac{(k+1)^5(2^k+3^k)}{k^5(2^{k+1} + 3^{k+1})}$$
Without using L'Hôpital's rule 5000 times?
Thanks!

It is a product of the following two expressions
$\frac{(k+1)^5}{k^5}=\left(1+\frac1k\right)^5$
$\frac{2^k+3^k}{2^{k+1}+3^{k+1}}= \frac{3^k(1+(2/3)^k)}{3^{k+1}(1+(2/3)^{k+1})}= \frac13\cdot$ $\frac{1+(2/3)^k}{1+(2/3)^{k+1}}$
