https://file.aiquickdraw.com/mathbot/file/1725397185523j7fbr7f6.jpg

The problem involves finding the limit of the function \( g(x) = \frac{2 \cos x^2 - 1}{\cos x - \sin x} \) as \( x \) approaches \(\frac{\pi}{4}\).

Let's go through the steps to solve this:

### Step 1: Evaluate the function at \( x = \frac{\pi}{4} \)
First, substitute \( x = \frac{\pi}{4} \) into the function:
\[
\cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
\]
\[
g\left(\frac{\pi}{4}\right) = \frac{2 \cos \left(\frac{\pi}{4}\right)^2 - 1}{\cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{4}\right)}
\]
Substitute \( \cos \left(\frac{\pi}{4}\right)^2 = \frac{1}{2} \):
\[
g\left(\frac{\pi}{4}\right) = \frac{2 \cdot \frac{1}{2} - 1}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{1 - 1}{0} = \frac{0}{0}
\]
This is an indeterminate form, so we need to apply **L'Hôpital's Rule**.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule can be applied because we have a \(\frac{0}{0}\) form. To do this, we differentiate the numerator and the denominator separately.

- **Numerator**: \( f(x) = 2\cos(x^2) - 1 \)
  - Differentiate \( f(x) \) using the chain rule:
  \[
  f'(x) = \frac{d}{dx} [2\cos(x^2) - 1] = 2 \cdot (-\sin(x^2)) \cdot 2x = -4x\sin(x^2)
  \]

- **Denominator**: \( h(x) = \cos(x) - \sin(x) \)
  - Differentiate \( h(x) \):
  \[
  h'(x) = -\sin(x) - \cos(x)
  \]

### Step 3: Re-evaluate the Limit
Now, we can re-evaluate the limit:
\[
\lim_{x \to \frac{\pi}{4}} g(x) = \lim_{x \to \frac{\pi}{4}} \frac{-4x \sin(x^2)}{-\sin(x) - \cos(x)}
\]

Substitute \( x = \frac{\pi}{4} \) into the derivatives:
- At \( x = \frac{\pi}{4} \), \( \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).
- Therefore, the limit becomes:
\[
\lim_{x \to \frac{\pi}{4}} \frac{-4 \cdot \frac{\pi}{4} \cdot \sin\left(\frac{\pi^2}{16}\right)}{-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \lim_{x \to \frac{\pi}{4}} \frac{-\pi \cdot \sin\left(\frac{\pi^2}{16}\right)}{-\sqrt{2}} = \frac{\pi \cdot \sin\left(\frac{\pi^2}{16}\right)}{\sqrt{2}}
\]

This expression doesn’t directly match any of the choices, but we can approximate or analyze further to conclude which of the answers is correct by looking at each option more closely.

### Conclusion
Given the nature of the problem and the form of the answers, option **A**: \(\lim_{x \to \frac{\pi}{4}} (2\cos x^2 - 1)\) is likely equivalent to the limit in question.

**Answer:** **A**

---

Would you like a more detailed explanation or do you have any other questions?

Here are 5 related questions for further practice:
1. How would you apply L'Hôpital's Rule if the denominator involved a trigonometric identity?
2. What is the limit of \(\frac{\sin(x)}{x}\) as \( x \) approaches 0?
3. How can you use Taylor series to evaluate limits?
4. How do you evaluate the limit of a piecewise function at a point of discontinuity?
5. What are the key conditions under which L'Hôpital's Rule can be applied?

**Tip:** When using L'Hôpital's Rule, always ensure the limit yields an indeterminate form before applying the rule.

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

Explore how to evaluate the limit of the function g(x) = (2cosx^2 - 1) / (cos x - sin x) as x approaches π/4 using L'Hôpital's Rule. This problem demonstrates key calculus concepts and is suitable for advanced high school or college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation