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Ok, one more...limit as x approaches 0 of (e^x+x)^1/x?
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[x]' = 1 ... so thats covered :)
err... is the exponent ^(1/x)?
(1+0)^înf = 1
yes
http://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+of+%28e^x%2Bx%29^1%2Fx goes to e^2 if this is right
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lim_(x->0) (e^x+x) = 1
e^(x+x)^(1/x) = e^(2x)*(1/x) = e^2
yep, the e^2 is the right answer
e^2
Just slot some figures in...
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