Math Problem Statement

limx→0+ (1−cosx)^sinx

Solution

To evaluate limx0+(1cosx)sinx\lim_{x \to 0^+} (1 - \cos x)^{\sin x}, let us break it into manageable parts.

Step-by-step solution:

  1. Rewrite the expression: y=(1cosx)sinxy = (1 - \cos x)^{\sin x}

    Taking the natural logarithm on both sides: lny=sinxln(1cosx)\ln y = \sin x \cdot \ln(1 - \cos x)

    The limit now becomes: limx0+lny=limx0+sinxln(1cosx)\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} \sin x \cdot \ln(1 - \cos x)

  2. Analyze ln(1cosx)\ln(1 - \cos x): As x0+x \to 0^+, (1cosx)(1 - \cos x) approaches 00. Using the approximation for small xx: 1cosxx221 - \cos x \approx \frac{x^2}{2} Thus: ln(1cosx)ln(x22)=ln(x2)ln2=2lnxln2\ln(1 - \cos x) \approx \ln\left(\frac{x^2}{2}\right) = \ln(x^2) - \ln 2 = 2 \ln x - \ln 2

  3. Substitute back: Substituting this into lny\ln y, we have: lnysinx(2lnxln2)\ln y \approx \sin x \cdot \left(2 \ln x - \ln 2\right)

  4. Analyze sinx\sin x: As x0+x \to 0^+, sinxx\sin x \approx x. Substituting: lnyx(2lnxln2)=2xlnxxln2\ln y \approx x \cdot \left(2 \ln x - \ln 2\right) = 2x \ln x - x \ln 2

  5. Evaluate each term:

    • For 2xlnx2x \ln x: limx0+2xlnx=limx0+2lnx1/x\lim_{x \to 0^+} 2x \ln x = \lim_{x \to 0^+} \frac{2 \ln x}{1/x} Using L’Hôpital’s Rule: =limx0+2/x1/x2=limx0+2x=0= \lim_{x \to 0^+} \frac{2/x}{-1/x^2} = \lim_{x \to 0^+} -2x = 0

    • For xln2-x \ln 2: limx0+xln2=0\lim_{x \to 0^+} -x \ln 2 = 0

    Thus, lny0\ln y \to 0.

  6. Exponentiate to get the result: y=elnye0=1y = e^{\ln y} \to e^0 = 1

Final Answer:

limx0+(1cosx)sinx=1\lim_{x \to 0^+} (1 - \cos x)^{\sin x} = 1


Would you like a deeper explanation of any step?

Here are five related questions to expand your understanding:

  1. How is L’Hôpital’s Rule applied in limits involving indeterminate forms?
  2. Why is (1cosx)x22(1 - \cos x) \approx \frac{x^2}{2} a good approximation for small xx?
  3. Can you explain why ln(x)\ln(x) dominates as x0+x \to 0^+?
  4. What happens if the base in this limit were different, e.g., (1cosx)x(1 - \cos x)^x?
  5. How does Taylor expansion help in solving limits involving trigonometric functions?

Tip: For exponential and logarithmic forms in limits, consider logarithmic properties to simplify calculations.

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Math Problem Analysis

Mathematical Concepts

Limits
Exponential and Logarithmic Functions
Trigonometric Approximations
Indeterminate Forms

Formulas

ln(a^b) = b * ln(a)
1 - cos(x) ≈ x^2 / 2 for small x
L'Hôpital's Rule for indeterminate forms of type 0/0

Theorems

L'Hôpital's Rule
Taylor Expansion for Trigonometric Functions
Natural Logarithm Properties

Suitable Grade Level

Undergraduate Calculus or Advanced High School Calculus