# Limit involving L'Hopital's Rule

## Homework Statement

$\displaystyle\lim_{x\rightarrow 0^+} x^{1/x}$

## The Attempt at a Solution

$y = \displaystyle\lim_{x\rightarrow 0^+} x^{1/x}$

$lny = \displaystyle\lim_{x\rightarrow 0^+} \frac {lnx}{x}$

This gives an indeterminate form and it's a quotient, so I can apply L'Hopital's Rule:

$\displaystyle\lim_{x\rightarrow 0^+} \frac{1}{x}$

As x approaches zero, the expression will grow towards infinity...

$lny = ∞$

$e^∞ = y$

$y = ∞$

I can't see anything wrong with what I'm doing, yet putting the limit into WolframAlpha spits out a 0. What am I doing wrong?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

$\displaystyle\lim_{x\rightarrow 0^+} x^{1/x}$

## The Attempt at a Solution

$y = \displaystyle\lim_{x\rightarrow 0^+} x^{1/x}$

$lny = \displaystyle\lim_{x\rightarrow 0^+} \frac {lnx}{x}$

This gives an indeterminate form and it's a quotient, so I can apply L'Hopital's Rule:

$\displaystyle\lim_{x\rightarrow 0^+} \frac{1}{x}$

As x approaches zero, the expression will grow towards infinity...

$lny = ∞$

$e^∞ = y$

$y = ∞$

I can't see anything wrong with what I'm doing, yet putting the limit into WolframAlpha spits out a 0. What am I doing wrong?

You do not have an indeterminate quotient, so you should not use l'Hospital's rule. In your case both factors 1/x and log(x) tend to ∞ in magnitude. However, be careful about the sign of log(x) for small x.

RGV

You do not have an indeterminate quotient, so you should not use l'Hospital's rule. In your case both factors 1/x and log(x) tend to ∞ in magnitude. However, be careful about the sign of log(x) for small x.

RGV

Sorry, I don't quite understand.

As x approaches 0, wouldn't the numerator go towards negative infinity and the denominator towards 0? This gives me (-∞/0) which I don't quite know how to interpret... That isn't considered an indeterminate form?

Where are my seeing this wrong?

Ray Vickson
Homework Helper
Dearly Missed
Sorry, I don't quite understand.

As x approaches 0, wouldn't the numerator go towards negative infinity and the denominator towards 0? This gives me (∞/0) which I don't quite know how to interpret...

Where are my seeing this wrong?

I am not allowed to go step-by-step through the argument; you just have to think about it more carefully.

RGV

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

$\displaystyle\lim_{x\rightarrow 0^+} x^{1/x}$

## The Attempt at a Solution

$y = \displaystyle\lim_{x\rightarrow 0^+} x^{1/x}$

$lny = \displaystyle\lim_{x\rightarrow 0^+} \frac {lnx}{x}$

This gives an indeterminate form and it's a quotient, so I can apply L'Hopital's Rule:
...
To be more explicit.

$\displaystyle \ln(y) = \lim_{x\to\,0^+} \frac {\ln(x)}{x}$ is not of an indeterminate form because:
$\displaystyle \ln(x)\to\,-\infty$

while $\displaystyle x\to\,0^+\ .$​
Use the fact that $\displaystyle \frac{1}{x}\to\,+\infty\ .$

Ah, okay. I guess I'm just seeing the problem wrong.

Thanks to both of you.

EDIT:

Ah, I was taking it farther than I needed to. The original expression as it is isn't an indeterminate form and may just be evaluated as is.

Last edited:
Ray Vickson
Homework Helper
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To be more explicit.

$\displaystyle \ln(y) = \lim_{x\to\,0^+} \frac {\ln(x)}{x}$ is not of an indeterminate form because:
$\displaystyle \ln(x)\to\,-\infty$

while $\displaystyle x\to\,0^+\ .$​
Use the fact that $\displaystyle \frac{1}{x}\to\,+\infty\ .$

I was hoping the OP would figure out that for himself; that is why I did not write it out in detail.

RGV

SammyS
Staff Emeritus
Homework Helper
Gold Member
I was hoping the OP would figure out that for himself; that is why I did not write it out in detail.

RGV
Ray,

I apologize for that.

I just interceded to point out to OP, exactly what line of his post you were commenting on, but I must have been too tired & got carried away.

Judging by his/her response, I'm still not convinced that OP understands why that expression is not indeterminate.

SammyS