# Basic limit question

1. Feb 25, 2014

### bobby2k

1. The problem statement, all variables and given/known data
Find the limit:

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

2. Relevant equations
L'Hopitals rule.

3. 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?

2. Feb 25, 2014

### Staff: Mentor

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

3. Feb 25, 2014

### 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.

4. Feb 27, 2014

### Staff: Mentor

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).

5. Feb 27, 2014

### scurty

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.

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