Proving the Limit of n(a1/n-1) for a > 0

[WrongMan]

Homework Statement

I have to prove that the limit as n ⇒ ∞ of: n(a^1/n-1) = log(a) -> For every a >0

Homework Equations

I have no idea what to use

The Attempt at a Solution

This was an exam question i left it blank, because i had (and have) no idea on where to even start, I've tried using integrals/antiderivatives, but i didnt seem to get anywhere, this part of the exam was about derivatives, antiderivateves, integrals and fundamental calculus theorems (which we use to plot functions, so i don't see this one applying here)
I know that if i set "a" as 1 that checks out, but that doesn't prove anything...
ive thought about using the left side of the equation as an exponent to "e" and setting it as equal to "a" but i don't see that helping.
What am i supposed to use to prove this?

 

[Mark44]

Homework Statement

I have to prove that the limit as n ⇒ ∞ of: n(a^1/n-1) = log(a) -> For every a >0

Homework Equations

I have no idea what to use

The Attempt at a Solution

This was an exam question i left it blank, because i had (and have) no idea on where to even start, I've tried using integrals/antiderivatives, but i didnt seem to get anywhere, this part of the exam was about derivatives, antiderivateves, integrals and fundamental calculus theorems (which we use to plot functions, so i don't see this one applying here)
I know that if i set "a" as 1 that checks out, but that doesn't prove anything...
ive thought about using the left side of the equation as an exponent to "e" and setting it as equal to "a" but i don't see that helping.
What am i supposed to use to prove this?

I would start by writing $n(a^{1/n} - 1)$ as $\frac{a^{1/n} - 1}{1/n}$ and then takiing the limit.

 

[WrongMan]

I would start by writing $n(a^{1/n} - 1)$ as $\frac{a^{1/n} - 1}{1/n}$ and then takiing the limit.

Hmm, i don't see it ...
as n becomes greater the nth root of a goes really near to 1, and if you take 1 out of it, you get a value between 0 and 1, divided by an (more or less) equally small number, it just leaves me at the same place as not doing that.

Should i take a derivative of both sides? that doesn't seem to help me either
i think i don't know which "limit solving algorithm" I am supposed to use

 

[Mark44]

Hmm, i don't see it ...
as n becomes greater the nth root of a goes really near to 1, and if you take 1 out of it, you get a value between 0 and 1, divided by an (more or less) equally small number, it just leaves me at the same place as not doing that.

So as $n \to \infty$, the numerator approaches 0, right? What does the denominator do?

Should i take a derivative of both sides? that doesn't seem to help me either
i think i don't know which "limit solving algorithm" I am supposed to use

 

[WrongMan]

So as $n \to \infty$, the numerator approaches 0, right? What does the denominator do?

also aproaches 0, because nth root goes to 1.
im not sure how to evaluate how fast they approach it and what the relation between it

 

[Mark44]

also aproaches 0, because nth root goes to 1.
im not sure how to evaluate how fast they approach it and what the relation between it

OK, so both numerator and denominator are approaching zero. There should be some rule (that's a hint) you've learned about that you can use.

 

[WrongMan]

OK, so both numerator and denominator are approaching zero. There should be some rule (that's a hint) you've learned about that you can use.

Riiiiight, i can only think of the l'hopital or cauchy's rule, i don't think that's it, but I'm going to think about that for a while, and try these and other methods, thanks, i'l upate here in a bit

 

[Mark44]

I was hinting at L'Hopital's Rule. Why do you think it doesn't apply here?

 

[WrongMan]

I was hinting at L'Hopital's Rule. Why do you think it doesn't apply here?

because that was one of the first things i thought of, it didnt seem like it was going to help as getting and exponential of((1-n)/n) didnt seem like a great simplification, I am trying this out now

 

[WrongMan]

so numerator look like 1/n * a^(1-n/n) and 1-n/n goes to -1.. so numerator = 1/n*a, this doesn't seem right ..

About the denominator, I should derivate in respect to n? wether i do this or not, the result is not correct.

am i supposed to derivate the nominator in respect to n?, i guess i should since its the unknown

 

[WrongMan]

Ok i got it right now.
Thanks for all the help :)

 

[Mark44]

so numerator look like 1/n * a^(1-n/n) and 1-n/n goes to -1.. so numerator = 1/n*a, this doesn't seem right ..

Correct, that's not right. $a^{1/n}$ is an exponential function, not a power function like $x^n$.

About the denominator, I should derivate in respect to n? wether i do this or not, the result is not correct.

"Derivate" is a word, but it's not one that's used in mathematics. You differentiate a function to get its derivative.

am i supposed to derivate the nominator in respect to n?, i guess i should since its the unknown

Ok i got it right now.
Thanks for all the help :)

Good deal. I'll mark this thread as Solved.

 

[WrongMan]

"Derivate" is a word, but it's not one that's used in mathematics. You differentiate a function to get its derivative.

Hahahah sorry for the bad english xD, i meant differentiate.