Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.lim x→0 (1 − 2x)1/x

To solve this limit, we can indeed use L'Hopital's Rule. L'Hopital's Rule states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives, provided the limit is in the form 0/0 or ∞/∞.

First, we rewrite the expression (1 - 2x)^(1/x) in the form of 0/0 or ∞/∞. We can do this by taking the natural logarithm of the expression, which gives us:

ln((1 - 2x)^(1/x)) = (1/x) * ln(1 - 2x)

Now, we can see that as x approaches 0, both the numerator and denominator approach 0, so we can apply L'Hopital's Rule.

Taking the derivative of the numerator, we get:

d/dx [ln(1 - 2x)] = -2 / (1 - 2x)

And the derivative of the denominator is:

d/dx [1/x] = -1 / x^2

So, applying L'Hopital's Rule, we get:

lim (x→0) [ln(1 - 2x) / (1/x)] = lim (x→0) [(-2 / (1 - 2x)) / (-1 / x^2)]

Simplifying the right side, we get:

= lim (x→0) [2x^2 / (1 - 2x)]

As x approaches 0, the limit is 0.

However, remember that we took the natural logarithm of the original expression, so to find the actual limit, we need to take the exponential of our answer:

e^0 = 1

So, the limit as x approaches 0 of (1 - 2x)^(1/x) is 1.