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L'Hopital's Rule

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Date: 8/11/96 at 17:55:17
From: Anonymous
Subject: Limit of x^(1/x)

Hi! I have a problem with a limit that can be solved with L'Hopital's
Rule because it is part of the first partial, but for which I don't
see how to apply L'Hopital's Rule:

Limit     x^(1/x)
x-> Inf

I am thankful for any help.
```

```
Date: 8/11/96 at 18:9:6
From: Doctor Paul
Subject: Re: Limit of x^(1/x)

Let's assume it is this instead:

Lt 1^(1/x)
x-> Inf

Note that as x goes to infinity, 1/x goes to zero, right?  so 1 to the
zero power is just 1.

I think it's pretty safe to assume that x^(1/x) also goes to 1 as x
goes to infinity.  Let's prove it:

You have:

Limit      x^(1/x)
x-> Inf

Let's say that's equal to sum number 'n'.  We want to solve for n...

Limit     x^(1/x)    =  n
x-> Inf

Take the natural log of both sides:

Limit     (1/x)*ln(x) = ln(n)
x-> Inf

Now we have to evaluate the left side. It's a limit that evaluates to
infinity/0.  Now we can use L'Hopital's Rule on the lefthand side.

Take the derivitive and re-evaluate the limit:

Limit    (1/x) / 1  = ln(n)
x-> Inf

Now we can see that the lefthand side evaluates to zero.

0 = ln(n)

Exponentiate both sides:

e^0 = n

so

n = 1

There's proof that the limit evaluates to one.

I hope this helps.

Regards,

-Doctor Paul,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus

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