Math Problem Statement

If g is the function defined by g(x) = (2cos^2(x) - 1) / (cos(x) - sin(x)), then lim(x → π/4) g(x) is equivalent to which of the following?

Solution

To solve this problem, we need to analyze the function given and evaluate the limit:

Problem Analysis

We have the function: g(x)=2cos2x1cosxsinxg(x) = \frac{2\cos^2 x - 1}{\cos x - \sin x} and we are asked to find limxπ4g(x)\lim_{x \to \frac{\pi}{4}} g(x).

Step 1: Simplify the Expression

We know that: 2cos2x1=cos(2x)2\cos^2 x - 1 = \cos(2x) Thus, we can rewrite g(x)g(x) as: g(x)=cos(2x)cosxsinxg(x) = \frac{\cos(2x)}{\cos x - \sin x}

Step 2: Substitute x=π4x = \frac{\pi}{4}

When x=π4x = \frac{\pi}{4}:

  • cos(π4)=sin(π4)=22\cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.
  • So cosxsinx=2222=0\cos x - \sin x = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0, making the denominator zero.
  • Similarly, 2x=π22x = \frac{\pi}{2}, so cos(2x)=cos(π2)=0\cos(2x) = \cos\left(\frac{\pi}{2}\right) = 0, making the numerator zero as well.

Since we have a 00\frac{0}{0} indeterminate form, we can apply L'Hôpital's Rule to evaluate the limit.

Step 3: Apply L'Hôpital's Rule

Differentiating the numerator and denominator with respect to xx:

  1. Numerator: The derivative of cos(2x)\cos(2x) is 2sin(2x)-2\sin(2x).
  2. Denominator: The derivative of cosxsinx\cos x - \sin x is sinxcosx-\sin x - \cos x.

So we get: limxπ4g(x)=limxπ42sin(2x)sinxcosx=limxπ42sin(2x)sinx+cosx\lim_{x \to \frac{\pi}{4}} g(x) = \lim_{x \to \frac{\pi}{4}} \frac{-2\sin(2x)}{-\sin x - \cos x} = \lim_{x \to \frac{\pi}{4}} \frac{2\sin(2x)}{\sin x + \cos x}

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

When x=π4x = \frac{\pi}{4}:

  • sin(2x)=sin(π2)=1\sin(2x) = \sin\left(\frac{\pi}{2}\right) = 1,
  • sinx+cosx=22+22=2\sin x + \cos x = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}.

Thus: limxπ4g(x)=212=22=2\lim_{x \to \frac{\pi}{4}} g(x) = \frac{2 \cdot 1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}

Conclusion

The correct answer is: 2\boxed{\sqrt{2}}

Looking at the answer choices, this matches Option B:

limxπ4(cosx+sinx)\lim_{x \to \frac{\pi}{4}} (\cos x + \sin x)

Let me know if you want details or have any questions.


Here are five related questions to expand your understanding:

  1. How does L'Hôpital's Rule help in evaluating limits involving indeterminate forms?
  2. What other trigonometric identities might be useful in simplifying expressions like 2cos2x12\cos^2 x - 1?
  3. How do we know when to apply L'Hôpital's Rule versus algebraic simplification?
  4. What is the geometric interpretation of limxπ4(cosx+sinx)\lim_{x \to \frac{\pi}{4}} (\cos x + \sin x)?
  5. How would the result change if the function had a different trigonometric expression in the denominator?

Tip: Remember to check for indeterminate forms before applying L'Hôpital's Rule.

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Identities
Indeterminate Forms
L'Hôpital's Rule

Formulas

lim(x → a) f(x) = L (limit definition)
cos(2x) = 2cos^2(x) - 1
L'Hôpital's Rule: lim(x → a) f(x) / g(x) = lim(x → a) f'(x) / g'(x)

Theorems

L'Hôpital's Rule
Trigonometric Identity for cos(2x)

Suitable Grade Level

Grades 11-12