Non-trivial limit

topsquark

Forum Staff
Jan 2006
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Wellsville, NY
How do you show that \(\displaystyle \lim_{n \to \infty} \left ( 1 + \frac{x}{n^2} \right ) ^n = 1\) ?

-Dan
 
Dec 2013
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757
Colombia
\(\displaystyle \lim_{n \to \infty} \left ( 1 + \frac{x}{n^2} \right ) ^n = \lim_{n \to \infty} \left ( 1 + \frac{x}{n^2} \right ) ^{\frac{n^2}{n}} = \lim_{n \to \infty} \left( \left ( 1 + \frac{x}{n^2} \right ) ^{n^2 }\right)^{\frac1n} =\left( e^x\right)0 = 1\)
 
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romsek

MHF Helper
Nov 2013
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California
\(\displaystyle \lim_{n \to \infty} \left ( 1 + \frac{x}{n^2} \right ) ^n = \lim_{n \to \infty} \left ( 1 + \frac{x}{n^2} \right ) ^{\frac{n^2}{n}} = \lim_{n \to \infty} \left( \left ( 1 + \frac{x}{n^2} \right ) ^{n^2 }\right)^{\frac1n} =\left( e^x\right)^0 = 1\)
small correction to set the final 0 as an exponent
 
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topsquark

Forum Staff
Jan 2006
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Wellsville, NY
Thanks to both of you. It was a lot easier than I thought! (Bow)

-Dan
 

Prove It

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Aug 2008
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How do you show that \(\displaystyle \lim_{n \to \infty} \left ( 1 + \frac{x}{n^2} \right ) ^n = 1\) ?

-Dan
$\displaystyle \begin{align*} \lim_{n \to \infty}{ \left[ \left( 1 + \frac{x}{n^2} \right) ^n \right] } &= \lim_{n \to \infty}{ \mathrm{e}^{ \ln{ \left[ \left( 1 + \frac{x}{n^2} \right) ^n \right] } } } \\ &= \lim_{n \to \infty}{ \mathrm{e}^{ n\ln{ \left( 1 + \frac{x}{n^2} \right) } } } \\ &= \lim_{n \to \infty}{ \mathrm{e}^{ \frac{\ln{ \left( 1 + \frac{x}{n^2} \right) }}{\frac{1}{n}} } } \\ &= \mathrm{e}^{ \lim_{n \to \infty}{ \frac{\ln{\left( 1 + \frac{x}{n^2} \right) }}{\frac{1}{n}} } } , \, \textrm{ a }\frac{0}{0}\,\textrm{indeterminate form} \\ &= \mathrm{e}^{ \lim_{n \to \infty}{ \frac{\ln{\left( n^2 + x \right) } - \ln{\left( n^2 \right) }}{\frac{1}{n}} } } \\ &= \mathrm{e}^{ \lim_{n \to \infty}{ \frac{ \frac{2\,n}{n^2 + x} - \frac{2}{n} }{-\frac{1}{n^2}} } } \textrm{ by L'Hospital's Rule} \\ &= \mathrm{e}^{ \lim_{n \to \infty}{ \left( 2\,n - \frac{2\,n^3}{n^2 + x} \right) } } \\ &= \mathrm{e}^{ \lim_{n \to \infty}{ \left( \frac{2\,n^3 + 2\,n\,x - 2\,n^3}{n^2 + x} \right) } } \\ &= \mathrm{e}^{ \lim_{n \to \infty}{ \frac{2\,n\,x}{n^2 + x} } } \\ &= \mathrm{e}^{ \lim_{n \to \infty}{ \frac{\frac{2\,x}{n}}{1 + \frac{x}{n^2 }} } } \\ &= \mathrm{e}^{ \frac{0}{1 + 0} } \\ &= \mathrm{e}^0 \\ &= 1 \end{align*}$
 
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