Math Problem Statement

lim (x -> 0) (1 - 2x)^(1/x)

Solution

To solve this limit using L'Hospital's Rule, we first need to express it in a form suitable for L'Hospital's Rule. The given limit is of an indeterminate form 11^\infty, so let's proceed with the following steps:

  1. Transform the Expression:

    Consider L=limx0(12x)1/x.L = \lim_{x \to 0} (1 - 2x)^{1/x}. Take the natural logarithm of both sides to bring the exponent down: lnL=limx0ln(12x)x.\ln L = \lim_{x \to 0} \frac{\ln(1 - 2x)}{x}.

  2. Apply L'Hospital's Rule:

    The form ln(12x)x\frac{\ln(1-2x)}{x} is of type 00\frac{0}{0} as x0x \to 0. Differentiate the numerator and denominator: Numerator: ddx[ln(12x)]=212x.\text{Numerator: } \frac{d}{dx}[\ln(1-2x)] = -\frac{2}{1-2x}. Denominator: ddx[x]=1.\text{Denominator: } \frac{d}{dx}[x] = 1.

    Apply L'Hospital's Rule: lnL=limx0212x.\ln L = \lim_{x \to 0} \frac{-2}{1-2x}.

  3. Evaluate the Limit:

    As x0x \to 0, lnL=210=2.\ln L = \frac{-2}{1 - 0} = -2.

  4. Exponentiate to Solve for LL:

    Since lnL=2\ln L = -2, then L=e2.L = e^{-2}.

So, the original limit is: limx0(12x)1/x=e2.\lim_{x \to 0} (1 - 2x)^{1/x} = e^{-2}.

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Math Problem Analysis

Mathematical Concepts

Limits
Exponential Functions
Logarithms
L'Hospital's Rule

Formulas

L'Hospital's Rule
ln(a^b) = b*ln(a)
lim (x -> 0) ln(1 - 2x)/x

Theorems

L'Hospital's Rule

Suitable Grade Level

Grades 11-12