I don't understand this statement from Wolfram Alpha:
Since $5^{2k+1}$ grows asymptotically slower than $3^{4k+1}$ as $k$ approaches $\infty$, $$\lim_{k\to\infty} 3^{-4k-1}\cdot 5^{2k+1} = 0.$$
Doesn't $5^x$ increase faster than $3^x$? Does it have something to do with
$$3^{-4k-1}\cdot 5^{2k+1}=\frac{5^{2k+1}}{3^{4k+1}}=\left(\frac{5}{3}\right)^{2k+1}\cdot\left(\frac{1}{3}\right)^{2k} \quad ?$$
You didn't pay attention to the exponents here; $$3^{4k+1}\text{ and }5^{2k+1}.$$
The left side can be simplified to get $3\cdot 3^{4k}=3\cdot 81^{k},$ while the right side is $5\cdot 5^{2k}=5\cdot 25^{k}.$ So we can observe that as $k$ increase linearly, the left side increases with a factor of $81,$ while the right side increases with a factor of $25.$
So that is the reason Wolfram Alpha said $3^{4k+1}$ grows faster than $5^{2k+1}.$ Hope this helps :)