# Hey anyone please can tell me this one:A=lim(e^(1/n*logn)) (n tends

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)

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)

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

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

hey the problem is

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

log n is not in numerator

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?

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