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

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

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    $\begingroup$ But $5^x$ grows more slowly than $3^{2x}=9^x$. $\endgroup$ May 14 at 4:08
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    $\begingroup$ And so $5^{2k+1} = 5 \cdot 25^k$ grows slower than $3^{4k+1}=3\cdot 81^k$. $\endgroup$ May 14 at 4:10
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    $\begingroup$ and $25^k$ is slower than $81^k$ $\endgroup$
    – Will Jagy
    May 14 at 4:11
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    $\begingroup$ 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. $\endgroup$ May 14 at 4:11
  • $\begingroup$ Thanks for the answers $\endgroup$
    – ianc1339
    May 14 at 4:23

1 Answer 1


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 :)


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