Question

Use L'Hopsital Rule
Lim x->0 (1-2x)^1/x

Answer 1

\[\lim_{x \rightarrow 0} (1-2x)^{1/x}\]

Answer 2

let \[\Large y = (1-2x)^{1/x}\]
take logs of both sides
\[\Large \ln y =\frac{ 1 }{ x } \ln (1-2x)\] put both sides to the power of e \[\huge y = e^{\frac{ \ln (1-2x) }{ x }}\] now put it back into your limit \[\huge \lim_{x \rightarrow 0} e^{\frac{ \ln (1-2x) }{ x }}\] gotta go, see if you can finish, or i'll be back later.

Answer 3

You can now apply l'hop to the exponent part of it.

Answer 4

Since if you plug in x=0, ln1/0 is 0/0.

Answer 5

cool, thankx Mr smith :)

Answer 6

Did you solve it? It should be e^(-2), after l'hopitals rule

wolfram agrees: http://www.wolframalpha.com/input/?i=%5Clim_%7Bx+%5Crightarrow+0%7D+%281-2x%29%5E%7B1%2Fx%7D