Math Problem Statement

Evaluate the following limit: \( \lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} \)

Solution

We are tasked with evaluating the following limit:

limx0cosxcosx3sin2x.\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}.

Let us break this problem into steps:

Step 1: Approximation of cosx\cos x

Near x=0x = 0, we use the Taylor series approximation: cosx1x22.\cos x \approx 1 - \frac{x^2}{2}.

Step 2: Expand cosx\sqrt{\cos x} and cosx3\sqrt[3]{\cos x}

Using binomial expansions:

  • For cosx=(cosx)1/2\sqrt{\cos x} = (\cos x)^{1/2}: cosx1x221x24.\sqrt{\cos x} \approx \sqrt{1 - \frac{x^2}{2}} \approx 1 - \frac{x^2}{4}.

  • For cosx3=(cosx)1/3\sqrt[3]{\cos x} = (\cos x)^{1/3}: [ \sqrt[3]{\cos x} \approx \sqrt[3]{1 - \frac{x^2}{2}} \approx 1 - \frac{x^2}{6}. ]

Step 3: Simplify the numerator

cosxcosx3(1x24)(1x26)=x26x24.\sqrt{\cos x} - \sqrt[3]{\cos x} \approx \left(1 - \frac{x^2}{4}\right) - \left(1 - \frac{x^2}{6}\right) = \frac{x^2}{6} - \frac{x^2}{4}.

Simplify further: cosxcosx3x212.\sqrt{\cos x} - \sqrt[3]{\cos x} \approx \frac{x^2}{12}.

Step 4: Approximation of sin2x\sin^2 x

Near x=0x = 0, sinxx\sin x \approx x. Thus: sin2xx2.\sin^2 x \approx x^2.

Step 5: Simplify the fraction

Substitute into the limit: cosxcosx3sin2xx212x2.\frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} \approx \frac{\frac{x^2}{12}}{x^2}.

Simplify: cosxcosx3sin2x112.\frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} \approx \frac{1}{12}.

Step 6: Evaluate the limit

As x0x \to 0, the limit is: limx0cosxcosx3sin2x=112.\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} = \frac{1}{12}.

Final Answer:

112\boxed{\frac{1}{12}}

Would you like a detailed explanation of any specific step? Here are some related questions to expand understanding:

  1. How do Taylor series expansions help in solving limits involving trigonometric functions?
  2. Can the same technique be applied to other similar problems involving higher-order roots?
  3. What is the justification for approximating sinxx\sin x \approx x near zero?
  4. How does the binomial theorem aid in expanding functions near their base values?
  5. Could L’Hôpital’s Rule be used to evaluate this limit instead?

Tip: When solving trigonometric limits, remember that Taylor approximations and simplifications near small angles (x0x \to 0) are often the most efficient methods

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
Taylor Series
Binomial Expansion

Formulas

cos x ≈ 1 - x^2/2
sqrt(cos x) ≈ 1 - x^2/4
cube root(cos x) ≈ 1 - x^2/6
sin^2 x ≈ x^2

Theorems

Taylor Series Approximation
Binomial Theorem

Suitable Grade Level

Grades 11-12