# Limit Marathon

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• Jun 9th 2008, 11:48 AM
Mathstud28
Quote:

Originally Posted by wingless
Since power series are restricted (by Mathstud) for this one, you can simply multiply $\sin x \ln x$ by $\frac{x}{x}$ like Krizalid did above.

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

but only if you show that $\lim_{x\to{c}}\frac{f(x)}{g(x)}=1$
• Jun 9th 2008, 12:35 PM
Mathstud28
Ok, here is a really really easy one.

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

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

EDIT: oops...a typo
• Jun 9th 2008, 01:28 PM
wingless
Quote:

Originally Posted by Mathstud28
Ok, here is a really really easy one.

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

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

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

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

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

$a$

---

A more formal approach is here:

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

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

Using L'hopital,

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

Now divide all terms by $a^x$, then all terms except $a^x \ln a$ and $a^x$ will go to zero. So $e^{ \frac{a^x \ln a}{a^x} } = e^{\ln a} = a$
• Jun 9th 2008, 01:33 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
Ok, here is a really really easy one.

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

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

EDIT: oops...a typo

No takers? Oh well,

There are a couple of ways to do this.

The smart*** way would be that since

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

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

This can be shown by the limit

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

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

Therefore

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

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

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

Now by termwise division with $a^x$

we get

$\lim_{x\to\infty}\frac{a+b\bigg(\frac{b}{a}\bigg)^ x+c\bigg(\frac{c}{a}\bigg)^x+d\bigg(\frac{d}{a}\bi gg)^x+...}{1+\bigg(\frac{b}{a}\bigg)^x+\bigg(\frac {c}{a}\bigg)^x+\bigg(\frac{d}{a}\bigg)^x+...}=a$
• Jun 9th 2008, 01:35 PM
Mathstud28
Quote:

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

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

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

$a$

---

A more formal approach is here:

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

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

Using $\color{red}{L'hopital's}$,

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

Now divide all terms by $a^x$, then all terms except $a^x \ln a$ and $a^x$ will go to zero. So $e^{ \frac{a^x \ln a}{a^x} } = e^{\ln a} = a$

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 PM
wingless
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 $\sim$, use the L'hopital solution.. =)
• Jun 9th 2008, 02:04 PM
Mathstud28
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 $\alpha$ such that

$\lim_{n\to\infty}\frac{(n!)^{\alpha}}{(3n)!}>1$
• Jun 9th 2008, 04:03 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
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 $\alpha$ such that

$\lim_{n\to\infty}\frac{(n!)^{\alpha}}{(3n)!}>1$

Here is a hint

So expanding we get

$\lim_{n\to\infty}\frac{(2\pi{n})^{\frac{\alpha}{2} }(n^n)^{\alpha}(e^{-n})^{\alpha}}{\sqrt{6\pi{n}}27^n(n^n)^3(e^{-n})^3}$
• Jun 9th 2008, 04:13 PM
topsquark
Quote:

Originally Posted by Mathstud28
Here is a hint

So expanding we get

$\lim_{n\to\infty}\frac{(2\pi{n})^{\frac{\alpha}{2} }(n^n)^{\alpha}(e^{-n})^{\alpha}}{\sqrt{6\pi{n}}27^n(n^n)^3(e^{-n})^3}$

Thank the Heaven's for Stirling's approximation. Some days it seems to me we wouldn't have Physics without it. :)

-Dan
• Jun 9th 2008, 04:43 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
Here is a hint

So expanding we get

$\lim_{n\to\infty}\frac{(2\pi{n})^{\frac{\alpha}{2} }(n^n)^{\alpha}(e^{-n})^{\alpha}}{\sqrt{6\pi{n}}27^n(n^n)^3(e^{-n})^3}$

Ok, since no one is answering.

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

Does no one have any hard limits?
• Jun 9th 2008, 04:49 PM
topsquark
Quote:

Originally Posted by Mathstud28
Ok, since no one is answering.

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

Not that you are incorrect in any way about this, but you are aware that this makes the limit go from 0 ( $\alpha = 3$) to $\infty$ ( $\alpha = 4$)?

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

-Dan
• Jun 9th 2008, 04:52 PM
Mathstud28
Quote:

Originally Posted by topsquark
Not that you are incorrect in any way about this, but you are aware that this makes the limit go from 0 ( $\alpha = 3$) to $\infty$ ( $\alpha = 4$)?

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

-Dan

I do not think so, for we would have to find a value such that a function of the form $a^n$ and a function $n^{b}$ are equal as n goes to infinity, which would be impossible since $n^b\prec{a^n}$
• Jun 9th 2008, 05:16 PM
galactus
Try these:

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

$\lim_{n\to\rightarrow{\infty}}\frac{1}{n}\sum_{j=1 }^{n}\sum_{k=1}^{n}\frac{j}{j^{2}+k^{2}}$
• Jun 9th 2008, 05:26 PM
Mathstud28
Quote:

Originally Posted by galactus
Try these:

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

$\lim_{n\to\rightarrow{\infty}}\frac{1}{n}\sum_{j=1 }^{n}\sum_{k=1}^{n}\frac{j}{j^{2}+k^{2}}$

Is the first one two?

$\lim_{y\to{0}}\frac{\int_0^{\pi}\tan(y\sin(x))dx}{ y}=\int_0^{\pi}\sin(x)dx=2$?
• Jun 9th 2008, 05:34 PM
topsquark
Quote:

Originally Posted by Mathstud28
Is the first one two?

$\lim_{y\to{0}}\frac{\int_0^{\pi}\tan(y\sin(x))dx}{ y}=\int_0^{\pi}\sin(x)dx=2$?

It would appear that Mathstud28 is doing another series expansion, but didn't tell us! (Rofl)

-Dan
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