Math Problem Statement

lim x-> 0 (1-(x+1) exp pow(x pow(-2)))/ x-0 such that f'(0) = 1

Solution

Let's break down the problem step by step.

The given limit expression is:

limx01(x+1)x2x\lim_{x \to 0} \frac{1 - (x+1)^{x^{-2}}}{x}

You also mention that f(0)=1f'(0) = 1, where f(x)f(x) is the function involved.

Step 1: Simplifying the expression

We need to evaluate this limit:

limx01(x+1)x2x\lim_{x \to 0} \frac{1 - (x+1)^{x^{-2}}}{x}

To handle the (x+1)x2(x+1)^{x^{-2}}, we will first take the logarithm of the expression inside the limit:

(x+1)x2=ex2ln(x+1)(x+1)^{x^{-2}} = e^{x^{-2} \ln(x+1)}

As x0x \to 0, we can approximate ln(x+1)\ln(x+1) using the Taylor series expansion around 0:

ln(x+1)xfor small values of x\ln(x+1) \approx x \quad \text{for small values of } x

Thus,

x2ln(x+1)x2x=1xx^{-2} \ln(x+1) \approx x^{-2} \cdot x = \frac{1}{x}

So,

(x+1)x2e1/x(x+1)^{x^{-2}} \approx e^{1/x}

Now the limit becomes:

limx01e1/xx\lim_{x \to 0} \frac{1 - e^{1/x}}{x}

Step 2: Evaluate the limit

Now we need to analyze the behavior of this expression as x0x \to 0.

  • As x0x \to 0, e1/xe^{1/x} \to \infty, which makes the numerator approach 1=1 - \infty = -\infty.
  • The denominator, xx, approaches 0.

Thus, the overall expression tends to 0\frac{-\infty}{0}, which suggests the limit might tend to -\infty, not 1. However, we are given that f(0)=1f'(0) = 1, so there might be a misunderstanding in either the problem setup or in the interpretation.

Step 3: Clarification and Next Steps

Given the information, it seems like either:

  1. The limit provided might need to be reevaluated, or
  2. There might be additional context (like a correction in the form of the function f(x)f(x)) that could change the outcome.

Would you like to clarify the original problem or check for potential adjustments?


Here are some follow-up questions to explore related concepts:

  1. What are the conditions for a limit to exist and be finite as x0x \to 0?
  2. How does the behavior of exponential functions affect limits?
  3. What is the Taylor series expansion of exe^x near zero?
  4. How can we determine the derivative of a function given its limit definition?
  5. How do we deal with limits that result in indeterminate forms like /0\infty/0?

Tip: Always check for potential approximations or series expansions when evaluating limits that involve exponential and logarithmic functions.

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Math Problem Analysis

Mathematical Concepts

Limits
Exponential Functions
Logarithmic Approximations
Differentiation

Formulas

lim x→0 (1 - (x+1)^(x^(-2))) / x
Taylor series approximation: ln(1 + x) ≈ x for small x
Exponential function property: a^b = e^(b ln a)

Theorems

L'Hôpital's Rule
Taylor Series Expansion
Limit Definition of Derivatives

Suitable Grade Level

Undergraduate Calculus