Finding the limit of e^(log(n)/n) as n approaches 0

[flash123]

hey anyone please can tell me this one:
A=lim(e^(1/n*logn)) (n tends to 0)
i took log on both sides and then by using l hospital rule
i arrive at lim(-1/n) (n tends to 0)
can't solve further...please help

 

[DonAntonio]

hey anyone please can tell me this one:
A=lim(e^(1/n*logn)) (n tends to 0)
i took log on both sides and then by using l hospital rule
i arrive at lim(-1/n) (n tends to 0)
can't solve further...please help

$$\lim_{x\to 0^+}\frac{\log x}{x}=-\infty\Longrightarrow \lim_{n\to 0}e^{\frac{\log n}{n}}=\lim_{x\to -\infty}e^x=0$$

Another way:

$$e^{\frac{\log n}{n}}=\left(e^{\log n}\right)^{1/n}=n^{1/n}\xrightarrow [n\to 0]{} 0$$

Of course, we assume in the above that $\,n\,$ is a continuous variable.

DonAntonio

 

[flash123]

hey the problem is

e^(1/(n*logn)) log n is with n

log n is not in numerator

 

[DonAntonio]

hey the problem is

e^(1/(n*logn)) log n is with n

log n is not in numerator

Yeat...too bad you didn't write parentheses in the OP to make that clear, uh?

 

[micromass]

This belongs in the homework forum and should contain an effort from the OP. Locked.