Indeterminate Products Giving Me Two Different Limits

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Homework Statement



lim x-->0 of x lnx

Homework Equations


The Attempt at a Solution



1) \frac{lnx}{1/x} = \frac{1/x}{-1/x^{2}} = (-x) = 0

2) \frac{x}{1/lnx} = \frac{1}{1/1/x} = \frac{1}{x} = \infty
 
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hmmm...are you sure \frac{d}{dx} \frac{1}{\ln (x)}=\frac{1}{\frac{1}{x}}?:wink:
 
yea I realized it after a minute of looking at it again. this is exactly the sort of stupid mistake that lowers my marks in tests.

and oops. I didn't see you answer or I wouldn't have edited the entry back.

edit edit: there, I put it back up... but for some reason I think I made a double of the thread... agh. it's 4 AM. I'm tired :smile:

edit edit edit: and, of course: thanks.

(I go sleep now)
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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