Limiting with L'Hospital's Rule: Evaluating (1-10x)^(1/x)

[davemoosehead]

Homework Statement

lim (1-10x)^(1/x)
x->0
evaluate the limit

Homework Equations

L'hostpital's rule

The Attempt at a Solution

take derivative:
lim (-10+100x)/x
x->0

can't divide by zero so take the derivative again but x goes away:
lim 100
x->0

is 100 the limit? is there a limit? now that I'm looking at it again, i don't think i have the derivative right...

 

[Sleek]

Take natural logs on both sides. You'll get,

ln(L) = lim(x->0) {1/x*ln(1-10x)} which is of the form 0/0. Apply l'Hospital's rule now. When done, convert the ln(L) = m {where m is the value of limit you got} again into exponential form, i.e. L=e^m.

Regards,
Sleek.

 

[HallsofIvy]

Homework Statement

lim (1-10x)^(1/x)
x->0
evaluate the limit

Homework Equations

L'hostpital's rule

The Attempt at a Solution

take derivative:
lim (-10+100x)/x
x->0

WHAT did you take the derivative of? I don't recognize that as having anything to do with your original limit!

can't divide by zero so take the derivative again but x goes away:
lim 100
x->0

is 100 the limit? is there a limit? now that I'm looking at it again, i don't think i have the derivative right...

Since your original form is NOT f(x)/g(x), the first thing I would do it take the logarithm:
If Y= $$(1-10x)^{1/x}$$ then ln(Y)= ln(1-10x)/x. Now apply L'Hopital's rule to that.

 

[davemoosehead]

ok...hows this look?

ln(L) = lim(x->0) { ln(1-10x)/x } = 0/0 so..

ln(L) = lim(x->0) { (-10)/(1-10x) }

ln(L) = -10

L = e^-10

 

[HallsofIvy]

Much better.

(Sorry about posting the same thing so many times. I got a bit carried away, didn't I?)