Find Limit of n*(x^(1/n)-1)-ln(x) for Any n

[bobby2k]

Homework Statement

Find the limit:

lim_{x -> \infty}[n*(x^{\frac{1}{n}}-1)-ln(x)], for any n.

Homework Equations

L'Hopitals rule.

The Attempt at a Solution

I know that the ratio of the first expression over the last goes to zero, by L'Hopital, but unfortunately I now have a difference and not a quotient. Is it possible to transform it in some way to use L'Hopital?

 

[mfb]

a-b = (ab-b^2)/b = (a/b - 1)/(1/b)
Not sure if one of those helps.

 

[scurty]

lim_{x -> \infty}[n*(x^{\frac{1}{n}}-1)-ln(x)], for any n.

$y = \displaystyle \lim_{x \to \infty}(ln[e^{n*(x^{\frac{1}{n}}-1)}]-ln(x))$
$y = \displaystyle \lim_{x \to \infty}\displaystyle ln\left[\frac{e^{n*(x^{\frac{1}{n}}-1)}}{x}\right]$
$e^y = \displaystyle \lim_{x \to \infty}\displaystyle \left[\frac{e^{n*(x^{\frac{1}{n}}-1)}}{x}\right]$

That gives it in fraction form. I'm not sure if it makes it easier to evaluate or not, I haven't tried the problem myself.

 

[mfb]

That gives it in fraction form. I'm not sure if it makes it easier to evaluate or not, I haven't tried the problem myself.

If you apply l'Hopital's rule on that fraction, you end up with the same fraction again (plus some irrelevant part you can ignore).

 

[scurty]

If you apply l'Hopital's rule on that fraction, you end up with the same fraction again (plus some irrelevant part you can ignore).

You're right.

I believe factoring out ln(x) so you have $\displaystyle\lim_{x\to\infty}\ln{(x)} \cdot \left(\frac{n(x^{1/n}-1)}{\ln{(x)}} - 1\right)$ might work. You can use l'Hospital's on the inside fraction which works better than what I suggested above.