Limit of Sequence: Find w/o L'Hosp. Rule

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SUMMARY

The limit of the sequence defined by \(\lim_{n \rightarrow \infty}\frac{n}{\log_{10}{n}}\) approaches \(+\infty\) as \(n\) increases. While L'Hospital's Rule provides a straightforward solution, alternative methods such as using subsequences or epsilon-delta proofs can also yield the same result. The discussion highlights the elegance of the subsequence approach, particularly the example \(\frac{10^{10^k}}{\log_{10}10^{10^k}}=10^{10^k-k}\), which clearly diverges to infinity. Participants emphasize the importance of understanding these alternative methods for a deeper grasp of limits.

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  • Knowledge of L'Hospital's Rule and its applications
  • Basic concepts of subsequences in mathematical analysis
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Homework Statement



Find the following limit:

\lim_{n \rightarrow \infty}\frac{n}{\log_{10}{n}}

The Attempt at a Solution



It's easy to find the limit using L'Hospital rule (after having used Heine theorem to transform the sequence into a function):

\lim_{x \rightarrow \infty}\frac{x}{\log_{10}{x}} = \lim_{x \rightarrow \infty}\frac{1}{\frac{1}{x\log{10}}} = +\infty

Is there any way of solving it without L'Hospital rule?

If I was to use the definition, then for every K, there should be such n_0 that for every n>n_0, (n/log_10(n)) > K. But I don't know how to solve this inequality. Any help would be greatly appreciated, thanks in advance!
 
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There are easily manageable subsequences like:
\frac{10^{10^k}}{\log_{10}10^{10^k}}=10^{10^k-k}
which clearly go to infinitely.

You could also work through an epsilon-delta proof of l'Hospital's rule.
 
The subsequence way of solving is elegant, i didn't realize it... thanks a lot!
 

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