Yes, but if anyone does any more, you can assert that $\displaystyle f(x)\sim{g(x)}\text{ as }x\to{c}$

but only if you show that $\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=1$

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- Jun 9th 2008, 11:48 AMMathstud28
- Jun 9th 2008, 12:35 PMMathstud28
Ok, here is a really really easy one.

$\displaystyle \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x+...\bigg)^{ \frac{1}{x}}$

where $\displaystyle a>b>c>d>...>0$

EDIT: oops...a typo - Jun 9th 2008, 01:28 PMwingless
$\displaystyle \lim_{x\to\infty}\left (a^x+b^x+c^x+d^x+...\right )^{\frac{1}{x}}$

$\displaystyle a^x$ will grow faster than all other terms. So we can ignore the others and conclude that,

$\displaystyle \lim_{x\to\infty}\left (a^x \right )^{\frac{1}{x}}$

$\displaystyle a$

---

A more formal approach is here:

$\displaystyle \lim_{x\to\infty}\left (a^x+b^x+c^x+d^x+...\right )^{\frac{1}{x}}$

$\displaystyle \lim_{x\to\infty} e^{ \frac{1}{x}\ln(a^x+b^x+c^x+d^x+...)}$

Using L'hopital,

$\displaystyle \lim_{x\to\infty} e^{ \frac{a^x \ln a + b^x \ln b + c^x \ln c + d^x \ln d + ...}{a^x + b^x + c^x + d^x + ...} }$

Now divide all terms by $\displaystyle a^x$, then all terms except $\displaystyle a^x \ln a$ and $\displaystyle a^x$ will go to zero. So $\displaystyle e^{ \frac{a^x \ln a}{a^x} } = e^{\ln a} = a$ - Jun 9th 2008, 01:33 PMMathstud28
No takers? Oh well,

There are a couple of ways to do this.

The smart*** way would be that since

$\displaystyle a>b>c>d>...>0$

we have that $\displaystyle a^x+b^x+c^x+d^x+...\sim{a^x}$

This can be shown by the limit

$\displaystyle \lim_{x\to\infty}\frac{a^x+b^x+c^x+d^x+...}{a^x}=\ lim_{x\to\infty}\frac{1+\bigg(\frac{b}{a}\bigg)^x+ \bigg(\frac{c}{a}\bigg)^x+\bigg(\frac{d}{a}\bigg)^ x+...}{1}=1$

this was gotten from the fact that $\displaystyle \lim_{x\to\infty}\bigg(\frac{\gamma}{\delta}\bigg) ^x=0$ if $\displaystyle \delta>\gamma$

Therefore

$\displaystyle \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x+...\bigg)^{ \frac{1}{x}}\sim\lim_{x\to\infty}(a^x)^{\frac{1}{x }}=a$

The other answer that is better in my opinion would be the use of the ratio/root test trick

$\displaystyle \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x\bigg)^{\fra c{1}{x}}=\lim_{x\to\infty}\frac{a\cdot{a^x}+b\cdot {b^x}+c\cdot{c^x}+d\cdot{d^x}+...}{a^x+b^x+c^x+d^x +...}$

Now by termwise division with $\displaystyle a^x$

we get

$\displaystyle \lim_{x\to\infty}\frac{a+b\bigg(\frac{b}{a}\bigg)^ x+c\bigg(\frac{c}{a}\bigg)^x+d\bigg(\frac{d}{a}\bi gg)^x+...}{1+\bigg(\frac{b}{a}\bigg)^x+\bigg(\frac {c}{a}\bigg)^x+\bigg(\frac{d}{a}\bigg)^x+...}=a$ - Jun 9th 2008, 01:35 PMMathstud28
Good answer! The only thing is, that you did not show that all the other terms can be ignored via a limit as I did in mine (No big deal, I think we all know it to be true)...and did you just use L'hopital's?:D

If anyone sees a problem with any of my solutions let me know - Jun 9th 2008, 01:45 PMwingless
Well, my first answer is the same as yours, I just didn't show the details. And if you don't want to evaluate the limit using $\displaystyle \sim$, use the L'hopital solution.. =)

- Jun 9th 2008, 02:04 PMMathstud28
EDIT:(sorry, I am making these up as we go) Tis is another easy one, if someone would like to post another one I would be more than happy to give up the posting position

Find the smallest integer$\displaystyle \alpha$ such that

$\displaystyle \lim_{n\to\infty}\frac{(n!)^{\alpha}}{(3n)!}>1$ - Jun 9th 2008, 04:03 PMMathstud28
- Jun 9th 2008, 04:13 PMtopsquark
- Jun 9th 2008, 04:43 PMMathstud28
Ok, since no one is answering.

It cannot be one or two, since that would leave a factor of $\displaystyle n^n$ and $\displaystyle \alpha=3$ would still leave that pesky $\displaystyle 27^n$ in the denominator, but $\displaystyle \alpha=4$ does the trick

Does no one have any hard limits? - Jun 9th 2008, 04:49 PMtopsquark
Not that you are incorrect in any way about this, but you are aware that this makes the limit go from 0 ($\displaystyle \alpha = 3$) to $\displaystyle \infty$ ($\displaystyle \alpha = 4$)?

I wonder if there is a way to find an $\displaystyle \alpha$ such that the limit is equal to 1?

-Dan - Jun 9th 2008, 04:52 PMMathstud28
- Jun 9th 2008, 05:16 PMgalactus
Try these:

$\displaystyle \lim_{y\to\rightarrow{0}}\frac{1}{y}\int_{0}^{\pi} tan(ysin(x))dx$

$\displaystyle \lim_{n\to\rightarrow{\infty}}\frac{1}{n}\sum_{j=1 }^{n}\sum_{k=1}^{n}\frac{j}{j^{2}+k^{2}}$ - Jun 9th 2008, 05:26 PMMathstud28
- Jun 9th 2008, 05:34 PMtopsquark