Solving a Long Limit Problem: Finding the Limit of e^x/x^n

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Hello,

I've been given a long limit problem to solve and i got stuck on this part in the question. Could someone please give me hints or suggestions on where to go next?

Given that \frac{e^{x}}{x^{n}} > e^{x - n\sqrt{x}}

Find the lim_{x\rightarrow+\infty} \frac{e^{x}}{x^{n}}



Well i know that lim_{x\rightarrow\infty} e^{x} = \infty

but i think the question would like us to use the inequality above.

Any help would be greatly appreciated :)
 
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I don't see how the inequality helps. Do you know about power series? If so, then can you write down a power series for the function e^x/x^n, and take its limit?
 
Tom Mattson said:
I don't see how the inequality helps. Do you know about power series? If so, then can you write down a power series for the function e^x/x^n, and take its limit?

umm... i haven't learned power series... Is there another approach i could take??
i tried l'hospitals rule but my answer didn't help much in this...
 
Hi caelestis! :smile:

Hint: prove that lim (x - n√x) = ∞. :smile:
 
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...

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