- #1

- 7

- 0

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)

cant solve further...please help

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- Thread starter flash123
- Start date

- #1

- 7

- 0

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)

cant solve further...please help

- #2

- 606

- 1

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)

cant 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 [itex]\,n\,[/itex] is a continuous variable.

DonAntonio

- #3

- 7

- 0

hey the problem is

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

log n is not in numerator

- #4

- 606

- 1

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?

- #5

- 22,129

- 3,298

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

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