Math Problem Statement

question c

Use power series to evaluate the limit: lim (x -> 0) ln(x + 1) / (sqrt(4 + x) - 2)

Solution

To solve these limits using power series, we can expand the functions in the numerator and denominator using their Taylor series at x=0x = 0 and then evaluate the limit based on the leading terms.

Let's go through each part one by one.

Part (a) limx01+xexx2\lim_{x \to 0} \frac{1 + x - e^x}{x^2}

  1. Expand exe^x in a Taylor series around x=0x = 0: ex=1+x+x22+x36+e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots

  2. Substitute this expansion into the expression: 1+xex=1+x(1+x+x22+x36+)1 + x - e^x = 1 + x - \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \right) Simplify by canceling terms: 1+xex=x22x361 + x - e^x = -\frac{x^2}{2} - \frac{x^3}{6} - \cdots

  3. Rewrite the limit: limx01+xexx2=limx0x22x36x2\lim_{x \to 0} \frac{1 + x - e^x}{x^2} = \lim_{x \to 0} \frac{-\frac{x^2}{2} - \frac{x^3}{6} - \cdots}{x^2}

  4. Divide each term by x2x^2: =limx0(12x6)= \lim_{x \to 0} \left(-\frac{1}{2} - \frac{x}{6} - \cdots \right)

  5. Take the limit as x0x \to 0: =12= -\frac{1}{2}

So, the answer for part (a) is: limx01+xexx2=12\lim_{x \to 0} \frac{1 + x - e^x}{x^2} = -\frac{1}{2}


Part (b) limx0xex1\lim_{x \to 0} \frac{x}{e^x - 1}

  1. Expand exe^x as before: ex=1+x+x22+e^x = 1 + x + \frac{x^2}{2} + \cdots So, ex1=x+x22+e^x - 1 = x + \frac{x^2}{2} + \cdots

  2. Rewrite the limit: limx0xx+x22+\lim_{x \to 0} \frac{x}{x + \frac{x^2}{2} + \cdots}

  3. Factor xx out from the denominator: =limx0xx(1+x2+)=limx011+x2+= \lim_{x \to 0} \frac{x}{x \left(1 + \frac{x}{2} + \cdots \right)} = \lim_{x \to 0} \frac{1}{1 + \frac{x}{2} + \cdots}

  4. Take the limit as x0x \to 0: =11=1= \frac{1}{1} = 1

So, the answer for part (b) is: limx0xex1=1\lim_{x \to 0} \frac{x}{e^x - 1} = 1


Part (c) limx0ln(x+1)4+x2\lim_{x \to 0} \frac{\ln(x+1)}{\sqrt{4 + x} - 2}

  1. Expand ln(x+1)\ln(x+1) using its Taylor series at x=0x = 0: ln(x+1)=xx22+x33\ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots

  2. Expand 4+x\sqrt{4 + x} using the binomial series around x=0x = 0: 4+x=2+x4x264+\sqrt{4 + x} = 2 + \frac{x}{4} - \frac{x^2}{64} + \cdots Therefore, 4+x2=x4x264+\sqrt{4 + x} - 2 = \frac{x}{4} - \frac{x^2}{64} + \cdots

  3. Rewrite the limit: limx0xx22+x4x264+\lim_{x \to 0} \frac{x - \frac{x^2}{2} + \cdots}{\frac{x}{4} - \frac{x^2}{64} + \cdots}

  4. Factor xx from both the numerator and denominator: =limx0x(1x2+)x(14x64+)=limx01x2+14x64+= \lim_{x \to 0} \frac{x \left(1 - \frac{x}{2} + \cdots \right)}{x \left(\frac{1}{4} - \frac{x}{64} + \cdots \right)} = \lim_{x \to 0} \frac{1 - \frac{x}{2} + \cdots}{\frac{1}{4} - \frac{x}{64} + \cdots}

  5. Take the limit as x0x \to 0: =114=4= \frac{1}{\frac{1}{4}} = 4

So, the answer for part (c) is: limx0ln(x+1)4+x2=4\lim_{x \to 0} \frac{\ln(x+1)}{\sqrt{4 + x} - 2} = 4


Would you like further clarification on any of these parts?

Follow-Up Questions:

  1. How can power series approximations be applied in calculus beyond limit calculations?
  2. What are the conditions required for a power series to converge?
  3. How does the Taylor series differ from the Maclaurin series?
  4. Can power series be used to approximate values for non-polynomial functions?
  5. Why is factoring useful when simplifying expressions involving power series?

Tip:

Remember, in limit problems, look for ways to factor or cancel terms to simplify calculations, especially with power series.

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Math Problem Analysis

Mathematical Concepts

Limits
Power Series
Taylor Series

Formulas

ln(1 + x) ≈ x - x^2 / 2 + x^3 / 3 - ...
sqrt(4 + x) ≈ 2 + x/4 - x^2/64 + ...

Theorems

Taylor Series Expansion

Suitable Grade Level

Grades 11-12, College Calculus