Question 1082790
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You wrote:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow0}\left(X^x\right)]


Which is trivial and equal to 1 since I have to assume that *[tex \LARGE X] is a constant.


On the other hand, if what you really meant was:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow0}\left(x^x\right)]


which, coincidentally, has the same answer but the solution is far from trivial.


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ a^x\ =\ e^{ln(a^x)}\ =\ e^{x\ln(a)}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^x\ =\ e^{x\ln(x)}]


Use the chain rule for limits:


If *[tex \Large \lim_{u\rightarrow b}f(u)\ =\ L] and *[tex \Large \lim_{x\rightarrow a}g(x)\ =\ b] and *[tex \Large f(x)] is continuous at *[tex \Large x\ =\ b] then *[tex \Large \lim_{x\rightarrow a}f\left(g(x)\right)\ =\ L]


Let *[tex \Large g(x)\ =\ x\ln(x)] and *[tex \Large f(u)\ =\ e^u]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow0}\left(x\ln(x)\right)]


Indeterminate form.  Rewrite to a L'Hôpital form.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow0}\left(\frac{\ln(x)}{\frac{1}{x}\right)]


Apply  L'Hôpital 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow0}\left(\frac{\frac{1}{x}}{-\frac{1}{x^2}\right)\ =\ \lim_{x\rightarrow0}(-x)\ =\ 0]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{u\rightarrow0}\left(e^u\right)\ =\ e^0\ =\ 1]


So by the chain rule


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow0}\left(x^x\right)\ =\ 1]


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">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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