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Limit of x^x^x as x>0 from right 
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#1
Dec608, 08:28 PM

P: 7

Any ideas?
lim X^{XX} x> 0^{+} I know how to do x^x: lim X^{X} = lim e^{x * ln x} = e^{0} = 0 lim x * ln x = lim ln x / (1/x) = lim (1/x) / (1/x^{2}) = lim x = 0 


#2
Dec608, 10:08 PM

P: 179

Now x^x is the exponent and x is the base, so do exactly as you just explained.



#3
Dec608, 10:12 PM

P: 7

Could you elaborate more? I alredy tried this method and got stuck with
lim (ln x) * (x^x), i could not solve it 


#4
Dec608, 10:15 PM

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Limit of x^x^x as x>0 from right



#5
Dec608, 10:24 PM

P: 7

lim ln x * (x^x) = lim ln x * lim x ^ x = ( infinity) * 0 = 0 as X > 0^{+}
So you want to tell me that lim (X^x^x) = lim e^{x^x *ln x} = e^0 = 1? The limit suppose to be 0, .000001 ^ ( .000001 ^ .000001 ) = very very very small number (checked with calculator) 


#6
Dec608, 10:27 PM

P: 179

Edit: I see, nevermind. 


#7
Dec608, 10:31 PM

P: 7

lim ln x * x was solved by switching it to lim ln x / (1/x) and taking derivatives, with lim ln x * x/x it's not gonna work 


#8
Dec608, 10:39 PM

P: 101

hence ......limx^x=1 as x tends to 0 from the right And lim (ln x) * lim x^x = infinity multiplied by 1 and NOT by 0 


#9
Dec608, 10:43 PM

P: 7

[edited for content], in my solutions it was 1, but somehow I copied formula wrong and the whole time assumes lim x^x = 0. I spend way too much time studying for finals.... gotta take break



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