Solving $\lim_{n\rightarrow \o}\frac{1}{x}^x$ Problem

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SUMMARY

The limit problem $\lim_{x\rightarrow 0^+}\left(\frac{1}{x}\right)^x$ can be solved using L'Hôpital's rule. By letting \( y = \left(\frac{1}{x}\right)^x \), we take the natural logarithm to obtain \( \ln(y) = -x \ln(x) \). This expression can be rewritten as \( -\frac{\ln(x)}{\frac{1}{x}} \), allowing the application of L'Hôpital's rule. The final result of the limit is 1.

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Physicsisfun2005
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Not sure if there is and answer

[tex]\lim_{n\rightarrow \o}\frac{1}{x}^x[/tex]


i suck at latexing lol...liimit approaching zero from the right i think...and its (1/x)^x...the quantity to the x power.
...i don't think i can use l'hospital's rule...so how do i solve it?
 
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(I assume it is x going to 0, not n!)

Sure you can use L'Hopital's rule. This is a (infinity)0 form so let y= (1/x)x. ln(y)= x ln(1/x)= -xln(x) and you can write that as [itex]-\frac{ln x}{\frac{1}{x}}[/itex]. Apply L'Hopital's rule to that. Whatever you get for ln y, the limit of the original problem is the exponential of that.
 
i got 1 as my answer...is that right?
 
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