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

• bobby2k
In summary, to find the limit of lim_{x -> \infty}[n*(x^{\frac{1}{n}}-1)-ln(x)], for any n, one can use L'Hopital's rule and transform the expression into a fraction form by factoring out ln(x). This allows for easier application of L'Hopital's rule and results in the same fraction again, which can then be evaluated to find the limit.
bobby2k

Homework Statement

Find the limit:

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

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?

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

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

scurty said:
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).

mfb said:
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.

1. What is the significance of finding the limit of n*(x^(1/n)-1)-ln(x) for any n?

Finding the limit of this expression can help us understand the behavior of the function as x approaches infinity. It can also help us determine if the function is continuous and differentiable at x=0.

2. How do you approach finding the limit of n*(x^(1/n)-1)-ln(x) for any n?

One approach is to use L'Hopital's rule, which involves taking the derivative of the numerator and denominator separately and then evaluating the limit. Another approach is to use substitution and algebraic manipulation to simplify the expression before taking the limit.

3. Is there a specific value for n that we need to consider when finding the limit?

No, the limit can be found for any value of n. However, the behavior of the function may differ for different values of n.

4. Can we use a graph to visualize the limit of n*(x^(1/n)-1)-ln(x) for any n?

Yes, a graph can help us visualize the behavior of the function as x approaches infinity and can also help us determine if the limit exists.

5. Are there any special cases we need to consider when finding the limit of n*(x^(1/n)-1)-ln(x) for any n?

Yes, we need to consider the cases where n=0 and n=1 separately, as they can lead to different limits. Additionally, for n<0, the limit may not exist.

Replies
17
Views
1K
Replies
5
Views
695
Replies
34
Views
2K
Replies
8
Views
1K
Replies
15
Views
2K
Replies
6
Views
785
Replies
13
Views
1K
Replies
23
Views
1K
Replies
14
Views
1K
Replies
7
Views
703