Question 890090
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\ \left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^2}}]


Exponentiate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\ e^{\frac{1}{x^2}\ln\left(\frac{\tan(x)}{x}\right)}\ =\ \lim_{x\rightarrow 0}\ e^{\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}]


Using the fact that *[tex \Large e^x] is continuous at *[tex \Large x\ =\ 0] and if *[tex \Large f(x)] is continuous around *[tex \Large x\ =\ g(a)] and *[tex \Large g(x)] is continuous around *[tex \Large x\ =\ a], then *[tex \Large \lim_{x\right a}\,f(g(x))\ =\ f\left( \lim_{x\rightarrow a}\,g(x)\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\ e^{\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}}\ =\ e^{\lim_{x\right 0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}}]


Use a Taylor Series expansion at *[tex \Large x\ =\ 0] to compute *[tex \Large \lim_{x\right 0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right 0}\,\frac{1}{3}\ +\ \frac{7x^2}{90}\ +\ O\left(x^4\right)\ =\ \frac{1}{3}]


Hence,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{\lim_{x\right 0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}}\ =\ e^{\frac{1}{3}}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>