# Limit problem - [applying L'Hopitals law/rule]

1. Jul 25, 2013

### SirPlus

Hi,
Find the limit of the function x^(1/x) as x tends to zero

I had assigned y to x^(1/x) and took the natural logarithm on both sides but that had not given me a quotient in its indetermined form that is 0/0 or infinity / infinity

maybe there is another approach not l'Hopitals rule ..

Thanks,

2. Jul 25, 2013

### SteamKing

Staff Emeritus
If y = x^(1/x) and you took the log of y, what did you get?

3. Jul 25, 2013

### SirPlus

I got 1/0 ...

4. Jul 25, 2013

### dirk_mec1

L'Hopital can not be used here, use small numbers to see the trend (from the right side of zero that is).

5. Jul 25, 2013

### hilbert2

First show that if $x \in ]0,1[$ , then $x^{1/x} \in ]0,x[$. Then deduce that this implies the limit is zero.

6. Jul 25, 2013

### vanhees71

Sure, you can use de L'Hospital's rule for the log of the function
$$\ln(x^x)=x \ln x, \quad x>0.$$
The limit for $x \rightarrow 0^+$ is of the type "$0 \cdot \infty$". This you easily can transform into a limit of the type "$\infty/\infty$"!

7. Jul 25, 2013

### dirk_mec1

@vanhees: that's not the limit, right?

8. Jul 25, 2013

### vanhees71

If you have the limit of $\ln[f(x)]$, then you can use $f(x)=\exp\{ \ln[f(x)] \}$ and the continuity of the exponential function to get the original limit!

9. Jul 25, 2013

### hilbert2

^ But it's $x^{1/x}$, not $x^{x}$. You will get something like "$-\infty / 0$"

10. Jul 25, 2013

### dirk_mec1

Exactly my point.

11. Jul 25, 2013

### vanhees71

Argh. I misread this from the beginning. Then of course you have
$$\ln[f(x)]=\frac{\ln x}{x},$$
and you don't need de L'Hospital at all. As a real function defined for $x>0$ you can take the limit from above $x \rightarrow 0^+$ (giving $-\infty$) and then apply the argument with the continuity of the exponential function.

12. Jul 25, 2013

### SirPlus

I don't get your argument vahess71 - i understand that as the x values decrease to the right the numerator as function tends to (- infinity) however the dinomenator tends to zero, the limit does not exist at all?

13. Jul 25, 2013

### hilbert2

How do you define a limit? Are you familiar with the epsilon-delta definition?

In the case of this problem, you can show that if x is a positive number that is close to zero, then $x^{1/x}$ is even closer to zero and it's obvious that $x^{1/x} \rightarrow 0$ when $x \rightarrow 0$.

Try giving $x$ the sequence of values $x=10^{-1},10^{-2},10^{-3},\dots$ and see what are the corresponding values of $x^{1/x}$.

14. Jul 25, 2013

### dirk_mec1

No, the limit is not obvious, try, like I said earlier, to fill in some small number in ln(x)/x and analyze the trend.

15. Jul 25, 2013

### SirPlus

As an answer, would it be easier for me to rather tabulate values of x between zero and one for functions in numerator and denomenator to observe the trend?

perhaps maybe sketch the graph for support - i guess its difficult to find the limit arithmetically

16. Jul 25, 2013

### Zondrina

$y = x^{1/x}$
$ln(y) = \frac{ln(x)}{x}$
$lim( ln(y) ) = lim( \frac{ln(x)}{x} )$ ( I'm intending the limit to be $x→0^+$ )
$ln(y) = lim( \frac{ln(x)}{x} )$

It is important to note that the two sided limit does not exist in this case because $ln(x)$ is defined only for $x > 0$ in the reals.

Now, observe the nature of the expression as we inch closer and closer to zero from the right. The numerator $ln(x)$ will decrease rapidly towards $-∞$ as we tend to zero. While the denominator $x$ will also decrease, but will tend towards zero.

It is important to note that $x > ln(x)$ for all $x > 0$. So the denominator tends to zero faster than the numerator tends to $-∞$. This causes the expression to 'blow up' because you know any quantity divided by something very very small is something very very large.

All this info tells us that $lim( \frac{ln(x)}{x} ) → -∞$ as $x→0^+$. So we get :

$ln(y) = -∞$
$e^{ln(y)} = e^{-∞}$
$y = 0$

This tells us that $lim(x^{1/x}) = 0$.

17. Jul 25, 2013

### Ray Vickson

We have $\ln(y) \to - \infty$ as $x \to 0+.$ While you are technically correct that the limit does not exist, you need to go back and look at what is actually meant by all this. It means that for any given number $N > 0$ we can find $\delta > 0$ so that $\ln(y) < -N$ if $0 < x < \delta.$ Thus, $0 < y < e^{-N}$ for $0 < x < \delta$. So, given any $\epsilon > 0$ we can choose $N > 0$ so that $e^{-N} \leq \epsilon$, and then we an choose $\delta > 0$ so that for $x \in (0,\delta)$ we have $|y(x)| < \epsilon$. That means that we have $\lim_{x \to 0+} y(x) =0.$