Question 1113261
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{n\right\infty}\ \frac{2n\ +\ 1}{2^n}]


The indeterminate form is *[tex \Large\ \frac{\infty}{\infty}] so apply L'Hôpital


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \lim_{n\right\infty}\ \frac{\frac{d}{dn}(2n\ +\ 1)}{\frac{d}{dn}\(2^n\)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \lim_{n\right\infty}\ \frac{2}{2n\cdot\ln(2)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \lim_{n\right\infty}\ \frac{2^{1-n}}{ln(2)}]


The limit of a constant times a function is equal to the constant times the limit of the function, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \frac{1}{\ln(2)}\,\cdot\,\lim_{n\right\infty}\ \(2^1\,\cdot\,2^{-n}\)\ =\ \frac{2}{\ln(2)}\,\cdot\,\lim_{n\right\infty}\ 2^{-n}]


I don't know how rigorous you need to get with this.  I'm satisfied that the limit is zero, but if you want to be absolutely correct, you need to use the chain rule for limits so you ultimately end up with *[tex \Large \lim_{u\right -\infty}\ e^u\ =\ 0]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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