# Limits involving exponents

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

$$3^{-4k-1}\cdot 5^{2k+1}=\frac{5^{2k+1}}{3^{4k+1}}=\left(\frac{5}{3}\right)^{2k+1}\cdot\left(\frac{1}{3}\right)^{2k} \quad ?$$

• But $5^x$ grows more slowly than $3^{2x}=9^x$. May 14 at 4:08
• And so $5^{2k+1} = 5 \cdot 25^k$ grows slower than $3^{4k+1}=3\cdot 81^k$. May 14 at 4:10
• and $25^k$ is slower than $81^k$ May 14 at 4:11
• Please use MathJax to format your question. It greatly helps the question visible to various search engines and hence is crucial for maintaining this community as a repository of mathematical knowledge. For some basic information about writing mathematics at this site see, e.g., here, here, here and here. I edited your question this time, but you can give a read to the linked postings and do it by yourself next time. May 14 at 4:11
• Thanks for the answers May 14 at 4:23

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 :)