# Limit Marathon

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• Jun 10th 2008, 12:56 PM
wingless
Quote:

Originally Posted by Mathstud28
$\displaystyle \lim_{x\to 0} \frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)} }} {{x^2 }}$

Another solution:

$\displaystyle \lim_{x\to 0} \frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)} }} {{x^2 }}$

$\displaystyle \lim_{x\to 0} \frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)}}}{x^2}\cdot \frac{{\sqrt {\cos \left( {ax} \right)} + \sqrt {\cos \left( {bx} \right)}}} {{\sqrt {\cos \left( {ax} \right)} + \sqrt {\cos \left( {bx} \right)}}}$

$\displaystyle \lim_{x\to 0} \frac{\cos (ax) - \cos (bx)}{x^2 \left ( \sqrt{\cos (ax)} + \sqrt{\cos (bx)}\right )}$

By using a trig identity,

$\displaystyle \lim_{x\to 0} \frac{ -2 \sin \left ( \frac{a x}{2}-\frac{b x}{2} \right ) \sin \left ( \frac{a x}{2}+\frac{b x}{2} \right ) }{x^2 \left ( \sqrt{\cos (ax)} + \sqrt{\cos (bx)}\right )}$

$\displaystyle \lim_{x\to 0} \frac{-2}{\sqrt{\cos (ax)} + \sqrt{\cos (bx)}} \cdot \frac{\sin \left ( x \frac{a-b}{2} \right )}{x} \cdot \frac{\sin \left (x \frac{a +b}{2} \right )}{x}$

By evaluating all limits,

$\displaystyle \frac{b^2-a^2}{4}$
• Jun 10th 2008, 03:01 PM
albi
Quote:

Originally Posted by Mathstud28
Ok, here are two more

$\displaystyle \lim_{n\to\infty}\bigg(\frac{3^n+5^n}{2^n+5^n}\big g)^{n(\sqrt{n+1}-\sqrt{n})}$

$\displaystyle \bigg(\frac{3^n+5^n}{2^n+5^n}\bigg)^{n(\sqrt{n+1}-\sqrt{n})} \le \bigg(\frac{3^n}{5^n}+1\bigg)^{n(\sqrt{n+1}-\sqrt{n})} = \left[\bigg(\frac{3^n}{5^n}+1\bigg)^\frac{5^n}{3^n}\righ t]^{n\frac{3^n}{5^n}(\sqrt{n+1}-\sqrt{n})}$

So $\displaystyle \lim_{n\to\infty} \bigg(\frac{3^n}{5^n}+1\bigg)^{n(\sqrt{n+1}-\sqrt{n})} = \exp\left( \lim_{n\to\infty} n\frac{3^n}{5^n}(\sqrt{n+1}-\sqrt{n}) \right)$

Obviously: $\displaystyle \lim_{n\to\infty}n\frac{3^n}{5^n} = 0$ and $\displaystyle \lim_{n\to\infty}(\sqrt{n+1}-\sqrt{n}) = 0$

Hence: $\displaystyle \lim_{n\to\infty} \bigg(\frac{3^n}{5^n}+1\bigg)^{n(\sqrt{n+1}-\sqrt{n})} = 1$

We see that: $\displaystyle \bigg(\frac{3^n+5^n}{2^n+5^n}\bigg)^{n(\sqrt{n+1}-\sqrt{n})} \ge 1$

Thus: $\displaystyle \lim_{n\to\infty}\bigg(\frac{3^n+5^n}{2^n+5^n}\big g)^{n(\sqrt{n+1}-\sqrt{n})} = 1$

I have one limit which I find realy hard, I will post it in a moment (have to check it out)
• Jun 10th 2008, 03:39 PM
albi
Try this one (without using Stirling formula and L'Hospital):

$\displaystyle \lim_{n \rightarrow \infty} \frac{(n!)^2 2^{2n}}{\sqrt{n} (2n)!}$
• Jun 10th 2008, 03:56 PM
Mathstud28
Quote:

Originally Posted by albi
Try this one (without using Stirling formula and L'Hospital):

$\displaystyle \lim_{n \rightarrow \infty} \frac{(n!)^2 2^{2n}}{\sqrt{n} (2n)!}$

Hmm...I can see the answer two ways...but ironically those two ways are L'hopital's and Stirlings approximation (Rofl)...and I am thinking it will have to do with an integral transform...am Ion the right track?
• Jun 10th 2008, 04:30 PM
galactus
Take a look at $\displaystyle \int_{-\infty}^{\infty}e^{-x^{2}}dx$
• Jun 10th 2008, 05:01 PM
Mathstud28
Quote:

Originally Posted by galactus
Take a look at $\displaystyle \int_{-\infty}^{\infty}e^{-x^{2}}dx$

Well that would correspond

$\displaystyle \lim_{n\to\infty}\frac{(n!)^24^n}{\sqrt{n}(2n)!}=\ frac{(n^n)^2(e^{-n})^24^n(2\pi{n})}{2\sqrt{\pi}n(n^n)^2(e^{-n})^2}=\sqrt{\pi}$

and $\displaystyle \int_{-\infty}^{\infty}e^{-x^2}=2\int_0^{\infty}e^{-x^2}=2\cdot\frac{\sqrt{\pi}}{2}=\sqrt{\pi}$
• Jun 10th 2008, 06:17 PM
Mathstud28
Try this one without L'hopitals

$\displaystyle \lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ln(\sin(b\ xi))}$
• Jun 10th 2008, 08:18 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
Try this one without L'hopitals

$\displaystyle \lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ln(\sin(b\ xi))}$

Hint: Power series

(There might be an easier way to do this)
• Jun 10th 2008, 10:08 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
Try this one without L'hopitals

$\displaystyle \lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ln(\sin(b\ xi))}$

Here is my solution

$\displaystyle \ln(\sin(a\xi))=(sin(a\xi)-1)-\frac{(sin(a\xi)-1)^2}{2}+...\sim\sin(a\xi)-1$

and by similar logic

$\displaystyle \ln(\sin(b\xi))\sim\sin(b\xi)-1$

$\displaystyle \therefore\lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ ln(\sin(b\xi))}\sim\lim_{\xi\to{0}}\frac{\sin(a\xi )-1}{\sin(b\xi)-1}=\frac{0-1}{0-1}=1$
• Jun 11th 2008, 09:03 AM
albi
Quote:

Originally Posted by Mathstud28
Hmm...I can see the answer two ways...but ironically those two ways are L'hopital's and Stirlings approximation (Rofl)...and I am thinking it will have to do with an integral transform...am Ion the right track?

I don't know what you mean by 'integral transform'. But (relativly) simple solution for this limit i have found a while ago. I use some function usually defined by the integral. What is this function? You just have it in your signature :P
• Jun 11th 2008, 09:28 AM
PaulRS
Quote:

Originally Posted by Mathstud28
Here is my solution

$\displaystyle \ln(\sin(a\xi))=(sin(a\xi)-1)-\frac{(sin(a\xi)-1)^2}{2}+...\sim\sin(a\xi)-1$

Are you sure?

Note that $\displaystyle \ln(\sin(a\xi))\rightarrow{-\infty}$ while $\displaystyle \sin(a\xi)-1\rightarrow{-1}$

how can $\displaystyle \ln(\sin(a\xi))\sim\sin(a\xi)-1$ be true?
• Jun 11th 2008, 11:38 AM
Mathstud28
Quote:

Originally Posted by PaulRS
Are you sure?

Note that $\displaystyle \ln(\sin(a\xi))\rightarrow{-\infty}$ while $\displaystyle \sin(a\xi)-1\rightarrow{-1}$

how can $\displaystyle \ln(\sin(a\xi))\sim\sin(a\xi)-1$ be true?

Yeah, I realized this, and I was unsure when I posted it, but I thought "Hey if I am wrong someone will correct me" and here you did. So what would your solution be?

I am actually kind of stumped without L'hopital's. If it is something easy I am going to kick myself

Actually

$\displaystyle \frac{\ln(\sin(a\xi))}{\ln(\sin(b\xi))}=\ln\bigg(\ sin(a\xi)^{\frac{1}{\ln(\sin(a\xi))}}\bigg)=L\Righ tarrow$$\displaystyle e^{L}=\sin(a\xi)^{\frac{1}{\ln(\sin(b\xi)}} So then we have that If \displaystyle e^{L}=\lim_{\xi\to{0}}\sin(a\xi)^{\frac{1}{\ln(\si n(b\xi)}}\Rightarrow{e^{-L}=\sin(a\xi)^{\frac{1}{-\ln(\sin(b\xi)}}} Now letting \displaystyle \psi=-\ln(\sin(b\xi))\Rightarrow{\xi=\frac{\arcsin(e^{-\psi})}{b}} and as \displaystyle x\to{0} \displaystyle \psi\to\infty So we have \displaystyle \lim_{\psi\to\infty}\bigg[\sin\bigg(\frac{a\arcsin(e^{-\psi})}{b}\bigg)\bigg]^{\frac{1}{\psi}} Which by the root/ratio test trick we get this limit is equal to \displaystyle \lim_{\psi\to\infty}\frac{\sin\bigg(\frac{a\arcsin (e^{-\psi-1})}{b}\bigg)}{\sin\bigg(\frac{a\arcsin(e^{-\psi})}{b}\bigg)} And since it can be seen that as \displaystyle x\to\infty that \displaystyle \sin\bigg(\frac{a\arcsin(e^{-\psi-1})}{b}\bigg)\sim\frac{a\arcsin(e^{-\psi-1})}{b}\sim\frac{ae^{-\psi-1}}{b} and it can also be said that \displaystyle \sin\bigg(\frac{a\arcsin(e^{-\psi})}{b}\bigg)\sim\frac{ae^{-\psi}}{b} So we have that this whole mess above is equivalent to \displaystyle \lim_{\psi\to\infty}\frac{e^{-\psi1}}{e^{-\psi}}=\frac{1}{e} So now we have \displaystyle e^{-L}=\frac{1}{e}\Rightarrow{L=1} (Sun) 2(Sun)(Sun)(Sun)th post! • Jun 11th 2008, 12:32 PM Mathstud28 Quote: Originally Posted by PaulRS Just note that: \displaystyle \sin \left( {a \cdot x} \right) = a \cdot x \cdot \left[ {1 + o\left( 1 \right)} \right] \Rightarrow \ln \left[ {\sin \left( {a \cdot x} \right)} \right] = \ln \left( a \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right] when \displaystyle x\rightarrow{0^+} \displaystyle \mathop {\lim }\limits_{x \to 0^ + } \frac{{\ln \left[ {\sin \left( {a \cdot x} \right)} \right]}} {{\ln \left[ {\sin \left( {b \cdot x} \right)} \right]}} = \mathop {\lim }\limits_{x \to 0^ + } \frac{{\ln \left( a \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]}} {{\ln \left( b \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]}} And: \displaystyle \frac{{\ln \left( a \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]}} {{\ln \left( b \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]}} =$$\displaystyle \frac{{\tfrac{{\ln \left( a \right)}} {{\ln \left( x \right)}} + 1 + \tfrac{{\ln \left[ {1 + o\left( 1 \right)} \right]}} {{\ln \left( x \right)}}}} {{\tfrac{{\ln \left( b \right)}} {{\ln \left( x \right)}} + 1 + \tfrac{{\ln \left[ {1 + o\left( 1 \right)} \right]}} {{\ln \left( x \right)}}}} \to 1$

$\displaystyle \mathop {\lim }\limits_{x \to 0^ + } \frac{{\ln \left[ {\sin \left( {a \cdot x} \right)} \right]}} {{\ln \left[ {\sin \left( {b \cdot x} \right)} \right]}} = 1$

I thought of doing it your way, except I would have just said $\displaystyle \sin(a\xi)\sim{a\xi}$ and done simliarly to you, but I do that so much I wanted to think of a more difficult way....you can find such a complicated way above (Tongueout)
• Jun 11th 2008, 02:40 PM
Mathstud28
Here is another easy one

$\displaystyle \lim_{x\to{0}}x^{x^{x^{x^{x^{x^{x^x}}}}}}$
• Jun 11th 2008, 03:42 PM
Krizalid
Try this one: $\displaystyle \lim_{\alpha\to\infty}\frac1{\alpha^2}\int_0^\alph a\ln(1+e^{x})\,dx.$
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