Need help with math
Don't know what I'm doing wrong here...
lim (x -> 0+) x^(1/x)
y = lim (x -> 0+) x^(1/x)
ln(y) = lim (x -> 0+) (1/x)ln(x)
ln(y) = lim (x -> 0+) ln(x)/x
As x approaches 0 from the right, the numerator approaches -∞ and the denominator approaches 0... this gives the form of (-∞/0)
Applying L'Hopital's Rule:
ln(y) = lim (x -> 0+) 1/x
As x approaches 0, the expression approaches ∞, so:
ln(y) = ∞
e^∞ = y
y = ∞
But this is wrong. What am I doing wrong?
September 22, 2012
16 Comments • Newest first
As x approaches 0, the expression approaches ∞, so:
ln(y) = -∞
e^-∞ = y
y = -∞
That infinity should be a negative infinity. e^-infinity does equal 0.
You don't use L'Hospital's rule here, because your limit is [-Infinity/0].
[b]L'Hospital's Rule only applies for [0/0] or [Infinity/Infinity] limits. [/b]
In your case:
ln y = lim x->0+ {ln[x]}/x = {1/0}*{-Infinity} = {Infinity}*{-Infinity}={-Infinity}
ln y = {-Infinity}
....y=e^{-Infinity} = 0
....y=0
[quote=SunsetDews]Ah nice catch. You're right. 0^∞ isn't a special limit... I was taking it farther than needed
(Fairly sure e^∞ = ∞ though...)
Thanks[/quote]
Yeah it is, I was thinking of something else. But glad we figured that other part out
[quote=DrHye]I made an edit because I may be confusing that special case with infinity^0
I think the answer's 0. I'm just about positive 0^infinity = 0 and isn't a special case for limits
And looking at your work... e^∞ = 0 not ∞[/quote]
Ah nice catch. You're right. 0^∞ isn't a special limit... I was taking it farther than needed
(Fairly sure e^∞ = ∞ though...)
Thanks
[quote=SunsetDews]Yeah, the way I do these is usually use logarithms to bring down the exponent which will give me an indeterminate product
Then using the fact that a*b = a/(1/b), I just rewrite it to make the product a quotient and use l'hopital's rule... but in this case, the exponent goes directly to a quotient and I don't see how else I'd manipulate it. I tried rearranging the terms to make it x^(-1)/(1/ln(x)) and that didn't work out either..[/quote]
I made an edit because I may be confusing that special case with infinity^0
I think the answer's 0. I'm just about positive 0^infinity = 0 and isn't a special case for limits
Yeah guys I know how to use wolframalpha... the problem is I'll eventually need to show an actual procedure during an exam lol.
Well, just doing the old fashioned noob way of plugging in values to the right of 0, you find that the limit approaches 0 as x approaches 0 from the right.
[quote=DrHye]When you take the limit of the initial problem lim (x -> 0+) x^(1/x), it's 0^(infinity)
I believe this is one of the special cases of limits, and I remember there were some steps you had to take if you were given a form like the one we have... But I don't remember what to do[/quote]
Yeah, the way I do these is usually use logarithms to bring down the exponent which will give me an indeterminate product
Then using the fact that a*b = a/(1/b), I just rewrite it to make the product a quotient and use l'hopital's rule... but in this case, the exponent goes directly to a quotient and I don't see how else I'd manipulate it. I tried rearranging the terms to make it x^(-1)/(1/ln(x)) and that didn't work out either..
it's equal to 0!
[quote=SunsetDews]Yeah that's what's striking me as the possible error... but I don't know how else I'd manipulate the limit and I can't really interpret ∞/0.[/quote]
When you take the limit of the initial problem lim (x -> 0+) x^(1/x), it's 0^(infinity)
I believe this is one of the special cases of limits, and I remember there were some steps you had to take if you were given a form like the one we have... But I don't remember what to do. I could be wrong though
But technically, 0 to any nonzero power is 0 (if the power is 0 I believe it's undefined). Do you have the answer key?
[quote=DrHye]Are you allowed to use l'hopital's rule if it's -∞/0? I think it has to be 0/0 or ∞/∞[/quote]
Yeah that's what's striking me as the possible error... but I don't know how else I'd manipulate the limit and I can't really interpret ∞/0.
I don't learn this before.
Are you allowed to use l'hopital's rule if it's -∞/0? I think it has to be 0/0 or ∞/∞
[quote=angtena]lim answers can be numbers or +-∞[/quote]
Yeah... my answer was infinity, but it's wrong. The limit's supposed to be 0.
lim answers can be numbers or +-∞
[quote=Mhmmmm]Stopped reading after "lim"[/quote]
why even post if u arn't going to help the TS? Stop spamming.
@TS: let me find my old math notes. I might be able to help you