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

[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,

 

[SteamKing]

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

 

[SirPlus]

I got 1/0 ...

 

[dirk_mec1]

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

 

[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.

 

[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"!

 

[dirk_mec1]

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

 

[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!

 

[hilbert2]

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

 

[dirk_mec1]

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

Exactly my point.

 

[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.

 

[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?

 

[hilbert2]

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?

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 $x^{1/x}$.

 

[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.

 

[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

 

[STEMucator]

$y = x^{1/x}$
$ln(y) = \frac{ln(x)}{x}$
$lim( ln(y) ) = lim( \frac{ln(x)}{x} )$ ( 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 $lim(x^{1/x}) = 0$.

 

[Ray Vickson]

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?

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 $\lim_{x \to 0+} y(x) =0.$