I am trying to solve this limit:
lim
The first result is an indetermination of the kind \frac{\infty}{\infty} but here applying L'Hopital would be too long, I do not see any substitution by means of equivalent infinitesimals possible and simplifying the limit neither.
How can I solve it? The solution must be \infty. Thank you!
Answer
L'Hopital rule applied 200 times leads to
\lim_{x\to\infty}\frac{(\log 2)^{200} \,2^x}{200!}=\infty
Or you can take the logarithm of the limit
\log\lim_{x\to\infty}\frac{2^x}{x^{200}}=\lim_{x\to\infty}\left(x\log 2-200\log x\right)=\lim_{x\to\infty}x\left(\log 2-200\frac{\log x}{x}\right)=\infty
As the log of the limit is +\infty the limit is +\infty
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