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

  • Thread starter Thread starter dobry_den
  • Start date Start date
  • Tags Tags
    Limit Sequence
dobry_den
Messages
113
Reaction score
0

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!
 
Last edited:
Physics news on Phys.org
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!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top