Use L’Hospital’s rule to evaluate lim(1−2x)^(1/x). (You must use L’Hospital’s ruleto receive credit. Do not use any other method.)

Answer 1

Raymond had the right idea, and he used l'Hopital's rule after the transformation, but he made an arithmetic error at the end.

2/(2x-1) has a limit of 2/-1 = -2 as x goes to 0.

Then, as we took the log before starting, we now take the exponential.

Thus, the solution is e^-2.

If you know what the answer should be at the beginning, it makes it easy to check the work!

Answer 2

(1-2x)^(1/x)

you need a fraction to use L-Hopital's rule

(1-2x)^1/x = (1+2y)^-1/y = 1/(1+2y)^1/y where y=-x

derivative of numerator =0

if there's a limit then the denominator's derivative doesn't equal zero. or at least a 2nd or 3rd or higher derivative doesn't = 0. if the derivative of the denominator approaches any number other than 0, the limit is zero.

if you try a few simple integers f(x) decreases, as if approaching zero, as x gets increasingly larger

another method, probably much better is take the log of f(x) then find the limit of the log.

log of (1-2x)^1/x =(1/x)log(1-2x) = log(1-2x)/x

take the derivative of numerator and denominator

1/(1-2x)(-2)/1 = -2/(1-2x) = 2/(2x-1) take the limit as x approaches infinity, it's = 0