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I am trying to solve this limit: $$\lim_{x\to\infty} \frac{2^x}{x^{200}}$$

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!

Evoked
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    hi! i'm sorry but I think they're different since I'm looking for a method to compute the limit and not theoretically, as I saw in that question. But, you probably know more mathemathics so if you still think so, should I delete the question? – Evoked Nov 01 '17 at 18:36
  • okay, and sorry for not realising it before! @Typhon – Evoked Nov 01 '17 at 18:38
  • @Typhon: There are several answers there which don't use L'Hospital. And there's absolutely nothing which is more difficult for the general case than for this special case. And if you want special cases, have a look at the list of linked questions. – Hans Lundmark Nov 01 '17 at 19:05

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$$\lim_{x \to \infty} \frac{2^x}{x^{200}} = \lim_{x \to \infty} \frac{2^x}{2^{200\log_2(x)}} = \lim_{x \to \infty} 2^{x-200 \log_2(x)} = 2^{\lim_{x \to \infty}x-200 \log_2(x)}$$

That limit clearly goes to infinity, so the whole function does as well

John Lou
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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 $+\infty$

Raffaele
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$$ \lim_{x\to\infty} \frac{2^x}{x^{200}}=\lim_{x\to\infty} \frac{e^{x\ln2}}{x^{200}}\rightarrow \mbox{(L'Hopital 200 times)}=\lim_{x\to\infty} \frac{(\ln2)^{(200)}e^{x\ln2}}{200!}=\infty $$

Anonymous
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