1. 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 $\lim_{x\to{c}}\frac{f(x)}{g(x)}=1$

2. 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

3. 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 $e^{ \frac{a^x \ln a}{a^x} } = e^{\ln a} = a$

4. 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

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

5. 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?

If anyone sees a problem with any of my solutions let me know

6. 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.. =)

7. EDITsorry, 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

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

8. Originally Posted by Mathstud28
EDITsorry, 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

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

9. 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

10. 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?

11. 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

12. 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 $n^b\prec{a^n}$

13. Try these:

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

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

14. 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?

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

15. 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!

-Dan

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