Math Problem Statement

Find lim

x→0+ (1 - e^x)^(1/x)

Solution

We are tasked with finding the limit:

limx0+(1ex)1x\lim_{x \to 0^+} \left( 1 - e^x \right)^{\frac{1}{x}}

Step 1: Understanding the behavior of the function as x0+x \to 0^+

As xx approaches 0 from the right, let's first examine the behavior of exe^x:

  • ex1e^x \to 1 as x0+x \to 0^+.
  • Therefore, 1ex01 - e^x \to 0 as x0+x \to 0^+.

So, the expression (1ex)1/x(1 - e^x)^{1/x} takes the form of 000^0, which is an indeterminate form. Thus, we need to manipulate this limit into a solvable expression.

Step 2: Taking the natural logarithm

To simplify the expression, we take the natural logarithm of the function inside the limit:

Let:

y=(1ex)1xy = \left( 1 - e^x \right)^{\frac{1}{x}}

Taking the natural logarithm of both sides:

ln(y)=ln((1ex)1x)\ln(y) = \ln \left( \left( 1 - e^x \right)^{\frac{1}{x}} \right)

Using the property of logarithms, ln(ab)=bln(a)\ln(a^b) = b \ln(a), we get:

ln(y)=1xln(1ex)\ln(y) = \frac{1}{x} \ln \left( 1 - e^x \right)

Step 3: Analyzing the limit of ln(y)\ln(y)

Now, we need to evaluate:

limx0+ln(1ex)x\lim_{x \to 0^+} \frac{\ln(1 - e^x)}{x}

Since 1ex01 - e^x \to 0 as x0+x \to 0^+, we are dealing with a logarithm of a small number. To handle this, we can expand exe^x using its Taylor series around x=0x = 0:

ex=1+x+x22+O(x3)e^x = 1 + x + \frac{x^2}{2} + O(x^3)

Thus,

1ex=1(1+x+x22+O(x3))=xx22+O(x3)1 - e^x = 1 - \left( 1 + x + \frac{x^2}{2} + O(x^3) \right) = -x - \frac{x^2}{2} + O(x^3)

So for small xx, we have:

1exx1 - e^x \approx -x

Now, substitute this into the logarithmic expression:

ln(1ex)ln(x)\ln(1 - e^x) \approx \ln(-x)

For small xx, ln(x)=ln(x)+iπ\ln(-x) = \ln(x) + i\pi (the complex logarithm), but for real values, we focus on the asymptotic behavior of ln(x)\ln(x).

Thus, we approximate the limit:

ln(1ex)xln(x)x\frac{\ln(1 - e^x)}{x} \approx \frac{\ln(x)}{x}

Step 4: Evaluating the limit

Now we analyze the limit of ln(x)x\frac{\ln(x)}{x} as x0+x \to 0^+. It is known that:

limx0+ln(x)x=\lim_{x \to 0^+} \frac{\ln(x)}{x} = -\infty

Step 5: Returning to the original function

Since ln(y)\ln(y) \to -\infty, it follows that:

y0y \to 0

Thus, the original limit is:

limx0+(1ex)1x=0\lim_{x \to 0^+} \left( 1 - e^x \right)^{\frac{1}{x}} = 0

Final Answer:

0\boxed{0}


Would you like a more detailed explanation or clarification on any step? Here are some related questions you could explore:

  1. How do you handle limits involving indeterminate forms like 000^0?
  2. What is the Taylor series expansion of exe^x and how does it help in limits?
  3. Can you explore the behavior of limits involving logarithms as x0x \to 0?
  4. What are some common techniques for solving indeterminate limits?
  5. How do you differentiate and expand logarithmic functions around small values?

Tip: When encountering limits of the form 000^0, consider using logarithms to transform the problem into a more manageable form.

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Math Problem Analysis

Mathematical Concepts

Limits
Indeterminate Forms
Logarithmic Functions
Exponential Functions
Taylor Series

Formulas

lim x→0+ (1 - e^x)^(1/x)
ln(a^b) = b * ln(a)
lim x→0+ ln(x)/x = -∞

Theorems

Logarithmic Limit Theorem
Indeterminate Form Analysis
Taylor Series Expansion

Suitable Grade Level

Grades 11-12