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 Post subject: limit
PostPosted: Tue, 28 Sep 2010 13:13:28 UTC 
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[ (1+x)^(1/x) -e-(e)( x)/2]/x^2
limit x---0

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 Post subject: Re: limit
PostPosted: Tue, 28 Sep 2010 15:20:50 UTC 
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mun wrote:
[ (1+x)^(1/x) -e-(e)( x)/2]/x^2
limit x---0


The limit does not exist.

Perhaps you mean \displaystyle
\lim_{x\to 0}\frac{(1+x)^{1/x}-e+\frac{e}{2}x}{x^2}=\frac{11}{24}e?

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: limit
PostPosted: Tue, 28 Sep 2010 16:35:22 UTC 
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outermeasure wrote:
mun wrote:
[ (1+x)^(1/x) -e-(e)( x)/2]/x^2
limit x---0


The limit does not exist.

Perhaps you mean \displaystyle
\lim_{x\to 0}\frac{(1+x)^{1/x}-e+\frac{e}{2}x}{x^2}=\frac{11}{24}e?[/quote
yes sir your answer is right but how

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 Post subject: Re: limit
PostPosted: Tue, 28 Sep 2010 16:44:17 UTC 
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mun wrote:
outermeasure wrote:
mun wrote:
[ (1+x)^(1/x) -e-(e)( x)/2]/x^2
limit x---0


The limit does not exist.

Perhaps you mean \displaystyle
\lim_{x\to 0}\frac{(1+x)^{1/x}-e+\frac{e}{2}x}{x^2}=\frac{11}{24}e?

yes sir your answer is right but how


Taylor expand (1+x)^{1/x}=\exp(\frac{1}{x}\log(1+x)) about x=0.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: limit
PostPosted: Tue, 28 Sep 2010 17:11:06 UTC 
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sir can we do this question without using taylor expansion.

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 Post subject: Re: limit
PostPosted: Tue, 28 Sep 2010 23:05:05 UTC 
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mun wrote:
sir can we do this question without using taylor expansion.


Maybe, but I can promise the work would be very ugly!
I think you should stick to Outermeasure's suggestion...you should know how to Taylor expand the exponential function!!

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 Post subject: Re: limit
PostPosted: Wed, 29 Sep 2010 04:56:29 UTC 
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mun wrote:
sir can we do this question without using taylor expansion.


You can. Justify why (1+x)^{1/x} is sufficiently smooth and hence find the second derivative of (1+x)^{1/x} at x=0. Now use Cauchy MVT.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: limit
PostPosted: Fri, 18 May 2012 15:36:51 UTC 
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outermeasure wrote:
mun wrote:
sir can we do this question without using taylor expansion.


You can. Justify why (1+x)^{1/x} is sufficiently smooth and hence find the second derivative of (1+x)^{1/x} at x=0. Now use Cauchy MVT.

sir can you explain it further more as i don't know how to apply this theorem on limits.

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 Post subject: Re: limit
PostPosted: Fri, 18 May 2012 16:39:32 UTC 
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Using outermeasure Hint

First We will find expansion of \displaystyle \bf{(1+x)^{\frac{1}{x}}} at \bf{x=0}

Let\displaystyle \bf{y=\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}}

Taking \bf{\log}on both side, We Get

\displaystyle \bf{\ln(y)=\lim_{x\rightarrow 0}\frac{1}{x}.\ln(1+x)=\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+......}{x}}

\displaystyle \bf{\ln(y)=\lim_{x\rightarrow 0}=1-\frac{x}{2}+\frac{x^2}{3}}

\displaystyle \bf{y=e^{1-\frac{x}{2}}+\frac{x^2}{3}=e-ex+\frac{11e}{24}.x^2+.....}

Now We Calculate Limit

\displaystyle \bf{\lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+ex}{x^2}}

Using Expansion of \displaystyle \bf{(1+x)^{\frac{1}{x}}}

\displaystyle \bf{\lim_{x\rightarrow 0}\frac{e-ex+\frac{11e}{24}+...-e+ex}{x^2}}

So \displaystyle \bf{\lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+ex}{x^2}=\frac{11e}{24}}


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 Post subject: Re: limit
PostPosted: Fri, 18 May 2012 16:41:06 UTC 
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To Outermeasure

Justify why (1+x)^{1/x} is sufficiently smooth and hence find the second derivative of (1+x)^{1/x} at x=0. Now use Cauchy MVT.[/quote]

sir can you explain it further more as i don't know how to apply this theorem on limits.[/quote]

I did not understand the above (2) line.

Thanks


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