# Limit Marathon

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• Jun 9th 2008, 10:08 AM
Mathstud28
Limit Marathon
Hello everyone, since every other type of marathon has been done on this site and there has been a lull in the activity around here, I propose a limit marathon. Same concept as all the other; after someone completes a limit and gets the OK from the limit's poster, he/she posts their own limit. The restrictions are, no L'hopital's unless stated otherwise (yes jump for joy) and no graphical help. Also, it would be nice if the poster of the limit would provide their solution. For the first one, we will start off with something simple

Find $\displaystyle \lim_{x\to{0}}\frac{\sin(2\sin(3\sin(4\sin(5x))))) }{x}$
• Jun 9th 2008, 10:48 AM
Sean12345
I'll take a logical shot seen as though i have never done limits.

$\displaystyle \sin (0)~=~0~~,~~a\times0~=~0$

Thus the limit is 0?
• Jun 9th 2008, 10:49 AM
Mathstud28
Quote:

Originally Posted by Sean12345
I'll take a logical shot seen as though i have never done limits.

[tex]\sin (0)~=~0~~,~~a\times0~=~0[\math]

Thus the limit is 0?

The limit is not zero, sorry.

The reason being that, your direct substitution method would yield $\displaystyle \frac{0}{0}$ which is indeterminate
• Jun 9th 2008, 10:51 AM
wingless
It's obviously 120, using power series.

$\displaystyle \sin x = x - \frac{x^3}{3!}...$

$\displaystyle \sin 5x = 5x - \frac{(5x)^3}{3!}...$

$\displaystyle 4\sin 5x = 4 \left (5x - \frac{(5x)^3}{3!}... \right )$

$\displaystyle \sin (4\sin 5x) = \left [ 4 \left (5x - \frac{(5x)^3}{3!}... \right ) \right ] - \frac{\left [ 4 \left (5x - \frac{(5x)^3}{3!}... \right ) \right ]^3}{3!}...$

You can easily see the pattern here. Now we can conclude,

$\displaystyle \sin ( 2 \sin (3 \sin (4 \sin 5x))))) = 120x + P(x)$
and P(x)'s degree is greater than 1.

So $\displaystyle \lim_{x\to 0}\frac{120x + P(x)}{x} = 120$.

I could also show it using sigma notation and it would be more formal, but characters get too little so.. ;p
• Jun 9th 2008, 10:59 AM
Mathstud28
Quote:

Originally Posted by wingless
It's obviously 120, using power series.

$\displaystyle \sin x = x - \frac{x^3}{3!}...$

$\displaystyle \sin 5x = 5x - \frac{(5x)^3}{3!}...$

$\displaystyle 4\sin 5x = 4 \left (5x - \frac{(5x)^3}{3!}... \right )$

$\displaystyle \sin (4\sin 5x) = \left [ 4 \left (5x - \frac{(5x)^3}{3!}... \right ) \right ] - \frac{\left [ 4 \left (5x - \frac{(5x)^3}{3!}... \right ) \right ]^3}{3!}...$

You can easily see the pattern here. Now we can conclude,

$\displaystyle \sin ( 2 \sin (3 \sin (4 \sin 5x))))) = 120x + P(x)$
and P(x)'s degree is greater than 1.

So $\displaystyle \lim_{x\to 0}\frac{120x + P(x)}{x} = 120$.

I could also show it using sigma notation and it would be more formal, but characters get too little so.. ;p

My solution is as follows

$\displaystyle \lim_{x\to{0}}\frac{\sin(u(x))}{u(x)}\sim{u(x)}\te xt { iff }u(0)=0$

This can be shown by

$\displaystyle \lim_{x\to{0}}\frac{sin(u(x))}{u(x)}$

Let $\displaystyle \psi=u(x)$

and since as $\displaystyle u(x)\to{0}=0\Rightarrow{\psi\to{0}}$ by the above statement we have that

$\displaystyle \lim_{\psi\to{0}}\frac{\sin(\psi)}{\psi}=1$

So then we can see that the above asymptotic equivalence is true

So we can then see

$\displaystyle \sin(2(\sin(3\sin(4\sin(5x))))\sim{2\sin(3\sin(4\s in(5x)))}$$\displaystyle \sim{3!\sin(4\sin(5x))}\sim{4!\sin(5x)}\sim{5!x}$

$\displaystyle \therefore\lim_{x\to{0}}\frac{\sin(2\sin(3\sin(4\s in(5x))))}{x}\sim\lim_{x\to{0}}\frac{5!x}{x}=5!$
• Jun 9th 2008, 11:04 AM
wingless
Quote:

Originally Posted by Mathstud28
My solution is as follows
(...)

Ah, I was so sure that you solved this limit using Power Series, as you always do (Rofl)
• Jun 9th 2008, 11:11 AM
Mathstud28
Quote:

Originally Posted by wingless
Ah, I was so sure that you solved this limit using Power Series, as you always do (Rofl)

Haha, thats always a good assumption, but alas! This time I think it was not the easiest, or most beautful way. Would you care to post a limit, or shall I post another?
• Jun 9th 2008, 11:14 AM
wingless
I can't think of a hard limit at the moment. Go on and post another (Sun)
• Jun 9th 2008, 11:16 AM
Mathstud28
Quote:

Originally Posted by wingless
I can't think of a hard limit at the moment. Go on and post another (Sun)

Ok, this is one that isnt too hard, but I have figured out only one way of doing it that would fit the restrictions of this marathon

Calculuate

$\displaystyle \lim_{x\to{0}}x^{\sin(x)}$

EDIT: For this one no power series (Smirk)
• Jun 9th 2008, 11:30 AM
Krizalid
$\displaystyle x^{\sin x}=\exp \left[ \sin x\ln x \right]=\exp \left[ \frac{\sin x}{x}\cdot x\ln x \right].$

As $\displaystyle x\to0,\,\frac{\sin x}x$ and $\displaystyle x\ln x$ exist, and the conclusion follows.
• Jun 9th 2008, 11:38 AM
Mathstud28
Quote:

Originally Posted by Krizalid
$\displaystyle x^{\sin x}=\exp \left[ \sin x\ln x \right]=\exp \left[ \frac{\sin x}{x}\cdot x\ln x \right].$

As $\displaystyle x\to0,\,\frac{\sin x}x$ and $\displaystyle x\ln x$ exist, and the conclusion follows.

That is the one I had. I also came up with this one

$\displaystyle \lim_{x\to{0}}x^{\sin(x)}$

Let $\displaystyle \sin(x)=\frac{1}{\psi}$

so that as $\displaystyle x\to{0}\Rightarrow{\psi\to\infty}$

So we would then have that

$\displaystyle \lim_{x\to{0}}x^{\sin(x)}=\lim_{\psi\to\infty}\arc sin\bigg(\frac{1}{\psi}\bigg)^{\frac{1}{\psi}}$

Now using the Root test/Ratio test tric we see that

$\displaystyle \lim_{\psi\to\infty}\arcsin\bigg(\frac{1}{\psi}\bi gg)^{\frac{1}{\psi}}=\lim_{\psi\to\infty}\frac{\ar csin\bigg(\frac{1}{\psi+1}\bigg)}{\arcsin\bigg(\fr ac{1}{\psi}\bigg)}$

Now obviously $\displaystyle \frac{1}{\psi+1}\sim\frac{1}{\psi}$

$\displaystyle \therefore\lim_{\psi\to\infty}\frac{\arcsin\bigg(\ frac{1}{\psi+1}\bigg)}{\arcsin\bigg(\frac{1}{\psi} \bigg)}\sim\lim_{\psi\to\infty}\frac{\arcsin\bigg( \frac{1}{\psi}\bigg)}{\arcsin\bigg(\frac{1}{\psi}\ bigg)}=1$

$\displaystyle \therefore\lim_{x\to{0}}x^{\sin(x)}=1$

I think this is pretty (Sun)

Go ahead and post yours if you wish
• Jun 9th 2008, 11:40 AM
Moo
Quote:

Originally Posted by Mathstud28
Ok, this is one that isnt too hard, but I have figured out only one way of doing it that would fit the restrictions of this marathon

Calculuate

$\displaystyle \lim_{x\to{0}}x^{\sin(x)}$

Hmmm, let me see... (Rofl)

$\displaystyle =\lim_{x \to 0} e^{\sin x \ln x} \sim_0 e^{x \ln x}$

By the way, did you know that $\displaystyle \sin x \sim_0 x$ comes from a similar form of power series ? o.O

The limit of $\displaystyle x \ln x$ in 0 is 0, by increasing comparison.

Therefore, the limit is 1.

Edit : eh, I'm more late than I imagined...
• Jun 9th 2008, 11:42 AM
Mathstud28
Quote:

Originally Posted by Moo
Hmmm, let me see... (Rofl)

$\displaystyle =\lim_{x \to 0} e^{\sin x \ln x} \sim_0 e^{x \ln x}$

By the way, did you know that $\displaystyle \sin x \sim_0 x$ comes from a similar form of power series ? o.O

The limit of $\displaystyle x \ln x$ in 0 is 0, by increasing comparison.

Therefore, the limit is 1.

Good job! Are you refering to the fact that $\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\sim{x}$

Also, the increasing rates thing suffices, and this is not jibe against you for I am sure you know this. But for other observers, but I think it is more apparent to show it this way

$\displaystyle \lim_{x\to{0}}x\ln(x)$

Let $\displaystyle x=\frac{1}{\psi}$

then as $\displaystyle x\to{0}$ we have that $\displaystyle \psi\to\infty$

$\displaystyle \therefore\lim_{x\to{0}}x\ln(x)=\lim_{x\to\infty}\ frac{-\ln(\psi)}{\psi}$

Which then equals zero due to the fact that $\displaystyle \ln(\psi)\prec\psi$
• Jun 9th 2008, 11:45 AM
Moo
Quote:

Originally Posted by Mathstud28
Good job! Are you refering to the fact that $\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\sim{x}$

Oh actually, it's Taylor series... I never noticed because it's not the same spelling in French so it wasn't the same for me.
But yeah, this looks like it (we learn it with the o(x^n) notation)
• Jun 9th 2008, 11:45 AM
wingless
Quote:

Originally Posted by Mathstud28
Good job! Are you refering to the fact that $\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\sim{x}$

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