Math Problem Statement

If g is the function defined by g(x) = \frac{\cos x - \sin x}{1 - 2\sin^2 x}, what is \lim_{x \to \frac{\pi}{4}} g(x)?

Solution

To evaluate the limit, we are given the function:

g(x)=cosxsinx12sin2xg(x) = \frac{\cos x - \sin x}{1 - 2 \sin^2 x}

and need to find:

limxπ4g(x)\lim_{x \to \frac{\pi}{4}} g(x)

Step 1: Simplify the Expression

We start by substituting x=π4x = \frac{\pi}{4} directly to check for any indeterminate form:

  1. For the numerator: cosπ4sinπ4=2222=0\cos \frac{\pi}{4} - \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0.
  2. For the denominator: 12sin2π4=12(22)2=1212=11=01 - 2 \sin^2 \frac{\pi}{4} = 1 - 2 \cdot \left( \frac{\sqrt{2}}{2} \right)^2 = 1 - 2 \cdot \frac{1}{2} = 1 - 1 = 0.

This yields a 00\frac{0}{0} indeterminate form, so we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

To apply L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to xx.

  1. Differentiate the numerator: ddx(cosxsinx)=sinxcosx\frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x

  2. Differentiate the denominator: ddx(12sin2x)=4sinxcosx=2sin(2x)\frac{d}{dx}(1 - 2 \sin^2 x) = -4 \sin x \cos x = -2 \sin(2x)

After applying L'Hôpital's Rule, we have:

limxπ4g(x)=limxπ4sinxcosx2sin(2x)\lim_{x \to \frac{\pi}{4}} g(x) = \lim_{x \to \frac{\pi}{4}} \frac{-\sin x - \cos x}{-2 \sin(2x)}

Step 3: Substitute x=π4x = \frac{\pi}{4} Again

Now, substitute x=π4x = \frac{\pi}{4}:

  1. For the numerator: sinπ4cosπ4=2222=2-\sin \frac{\pi}{4} - \cos \frac{\pi}{4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}.
  2. For the denominator: 2sin(2π4)=2sin(π2)=21=2-2 \sin\left(2 \cdot \frac{\pi}{4}\right) = -2 \sin\left(\frac{\pi}{2}\right) = -2 \cdot 1 = -2.

Thus,

limxπ4g(x)=22=22\lim_{x \to \frac{\pi}{4}} g(x) = \frac{-\sqrt{2}}{-2} = \frac{\sqrt{2}}{2}

Final Answer

The correct answer is:

12\boxed{\frac{1}{\sqrt{2}}}

or, in decimal form, this is equivalent to answer choice B.

Would you like more details or have any questions?

Related Questions:

  1. How would we approach the limit if L'Hôpital's Rule didn't apply?
  2. What are common indeterminate forms that lead to using L'Hôpital's Rule?
  3. How do trigonometric identities help in simplifying limits?
  4. How can we use Taylor series expansions for similar trigonometric limits?
  5. What if we needed to evaluate limxπ2g(x)\lim_{x \to \frac{\pi}{2}} g(x)?

Tip:

When dealing with trigonometric limits, checking for direct substitution and possible simplifications using trigonometric identities can often save time before jumping to L'Hôpital's Rule.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Trigonometry
L'Hôpital's Rule
Indeterminate Forms

Formulas

L'Hôpital's Rule: \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} if \frac{f(c)}{g(c)} is indeterminate
Trigonometric Identities: \sin^2 x + \cos^2 x = 1

Theorems

L'Hôpital's Rule

Suitable Grade Level

Grades 11-12 or Introductory College Calculus