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Limit involving L'Hopital's Rule

  • Thread starter Mangoes
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  • #1
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Homework Statement



[itex]
\displaystyle\lim_{x\rightarrow 0^+} x^{1/x}
[/itex]



The Attempt at a Solution



[itex]
y = \displaystyle\lim_{x\rightarrow 0^+} x^{1/x}
[/itex]

[itex]
lny = \displaystyle\lim_{x\rightarrow 0^+} \frac {lnx}{x}
[/itex]

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

[itex]
\displaystyle\lim_{x\rightarrow 0^+} \frac{1}{x}
[/itex]

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

[itex]
lny = ∞
[/itex]

[itex]
e^∞ = y
[/itex]

[itex]
y = ∞
[/itex]


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?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



[itex]
\displaystyle\lim_{x\rightarrow 0^+} x^{1/x}
[/itex]



The Attempt at a Solution



[itex]
y = \displaystyle\lim_{x\rightarrow 0^+} x^{1/x}
[/itex]

[itex]
lny = \displaystyle\lim_{x\rightarrow 0^+} \frac {lnx}{x}
[/itex]

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

[itex]
\displaystyle\lim_{x\rightarrow 0^+} \frac{1}{x}
[/itex]

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

[itex]
lny = ∞
[/itex]

[itex]
e^∞ = y
[/itex]

[itex]
y = ∞
[/itex]


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
 
  • #3
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0
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?
 
  • #4
Ray Vickson
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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
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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Homework Statement



[itex]
\displaystyle\lim_{x\rightarrow 0^+} x^{1/x}
[/itex]

The Attempt at a Solution



[itex]
y = \displaystyle\lim_{x\rightarrow 0^+} x^{1/x}
[/itex]

[itex]
lny = \displaystyle\lim_{x\rightarrow 0^+} \frac {lnx}{x}
[/itex]

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

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

while [itex]\displaystyle x\to\,0^+\ .[/itex]​
Use the fact that [itex]\displaystyle \frac{1}{x}\to\,+\infty\ .[/itex]
 
  • #6
96
0
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:
  • #7
Ray Vickson
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To be more explicit.

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

while [itex]\displaystyle x\to\,0^+\ .[/itex]​
Use the fact that [itex]\displaystyle \frac{1}{x}\to\,+\infty\ .[/itex]
I was hoping the OP would figure out that for himself; that is why I did not write it out in detail.

RGV
 
  • #8
SammyS
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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
 

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