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I'm trying to prove $\lim_{x\rightarrow \infty}$$\frac{25x^2}{e^{10x}}=0 $?

Should I first consider $\lim_{x\rightarrow \infty}$$\frac{5x}{e^{5x}}$, and then assume $\lim_{x\rightarrow \infty}$$[\frac{5x}{e^{5x}}]^2$?

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    $\begingroup$ You can but one way or another, you are going to need that the exponential function dominates the polynomial: this is either something you know (as a standard limit) or you can use theorems like l'Hôpital's. $\endgroup$ – StackTD Dec 9 '16 at 11:58
  • $\begingroup$ what theorems are you allowed to use? $\endgroup$ – Siddharth Bhat Dec 9 '16 at 12:01
  • $\begingroup$ We're allowed to use L’Hospital’s Rule. Would I therefore consider $lim_{x\rightarrow \infty}\frac{g(x)}{f(x)}$ as $lim_{x\rightarrow \infty}\frac{g'(x)}{f'(x)}$ $\endgroup$ – Jeremy Tompson Dec 9 '16 at 12:05
  • $\begingroup$ Yes, you can apply l'Hôpital's rule twice. $\endgroup$ – StackTD Dec 9 '16 at 12:11
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You can use L'Hopital rule. It says that if $f,g$ are differentiable functions on $\mathbb{R}$, then $$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} $$ Using two steps of the above rule you get $$ \lim_{x \to \infty} \frac{25x^2}{e^{10x}} = \lim_{x \to \infty} \frac{50x}{10e^{10x}} = \lim_{x \to \infty} \frac{50}{100e^{10x}} = 0 $$ Let me explain what happened here: In the first equality we used that the derivative $25x^2 = 50x$, and the derivative of $e^{10x} = 10 e^{10x}$. Then in the second equality we used that the derivative of $50x$ is 50, and derivative of $10e^{10x} = 100e^{10x}$.

This got rid of the dependence of $x$ in the numerator, and brought the limit on a form you know how to deal with: $$ \lim_{x \to \infty} \frac{50}{100e^{10x}} = \frac{50}{\infty} = 0, $$ because the numerator is a constant for all $x$, whilst the denominator goes to infinity, and therefore the limit is zero.

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Yes, you can certainly note that the substitution $5x=t$ brings the limit into the form $$ \lim_{x\to\infty}\frac{25x^2}{e^{10x}}= \lim_{t\to\infty}\frac{t^2}{e^{2t}}= \lim_{t\to\infty}\left(\frac{t}{e^t}\right)^{\!2}=0 $$ because $$ \lim_{t\to\infty}\frac{t}{e^t}=0 $$ You can also do it without the substitution, reducing to $$ \lim_{x\to\infty}\frac{5x}{e^{5x}}=0 $$

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For $x>0$ we have $e^{10x} =1+10x+ \frac{100x^2}{2!}+\frac{1000x^3}{3!}+... > \frac{1000x^3}{3!}$. Then

$\frac{e^{10x}}{25x^2}> \frac{20x}{3}$. Thus

$0 \le \frac{25x^2}{e^{10x}}< \frac{3}{20x}$

and the result follows

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HINT.- Yes, you can. On the other hand you have $$\lim_{x\to\infty}\frac{P_m(x)}{P_n(x)}=0$$ where $m,n$ are the degrees of the polynomials and $m\lt n$ so the same limit if $P_n(x)$ is the exponential function.

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