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We are given the function:

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

and we are asked to find:

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

### Step 1: Substitute \( x = \frac{\pi}{4} \)
First, substitute \( x = \frac{\pi}{4} \) directly into the function to see if we encounter any indeterminate forms:

\[
g\left(\frac{\pi}{4}\right) = \frac{\cos{\frac{\pi}{4}} - \sin{\frac{\pi}{4}}}{1 - 2\sin^2{\frac{\pi}{4}}}
\]

We know:
\[
\cos{\frac{\pi}{4}} = \sin{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}
\]

Substituting this into the expression:

\[
g\left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{1 - 2\left(\frac{\sqrt{2}}{2}\right)^2}
\]

This simplifies to:

\[
g\left(\frac{\pi}{4}\right) = \frac{0}{1 - 2 \cdot \frac{1}{2}} = \frac{0}{1 - 1} = \frac{0}{0}
\]

So, we have an indeterminate form \(\frac{0}{0}\), which suggests we need to apply **L'Hopital's Rule**.

### Step 2: Apply L'Hopital's Rule
Since we have the indeterminate form \(\frac{0}{0}\), we can differentiate the numerator and the denominator separately and then evaluate the limit.

#### Differentiate the numerator:
\[
\frac{d}{dx} \left(\cos{x} - \sin{x}\right) = -\sin{x} - \cos{x}
\]

#### Differentiate the denominator:
\[
\frac{d}{dx} \left(1 - 2\sin^2{x}\right) = -4\sin{x}\cos{x}
\]

Now, the limit becomes:

\[
\lim_{x \to \frac{\pi}{4}} \frac{-\sin{x} - \cos{x}}{-4\sin{x}\cos{x}}
\]

### Step 3: Substitute \( x = \frac{\pi}{4} \) again
At \( x = \frac{\pi}{4} \), we know:

\[
\sin{\frac{\pi}{4}} = \cos{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}
\]

Substitute these values into the new expression:

\[
\lim_{x \to \frac{\pi}{4}} \frac{-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{-4 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}}
\]

Simplifying:

\[
= \frac{-\sqrt{2}}{-4 \cdot \frac{1}{2}} = \frac{-\sqrt{2}}{-2} = \frac{\sqrt{2}}{2}
\]

### Final Answer:
\[
\lim_{x \to \frac{\pi}{4}} g(x) = \frac{\sqrt{2}}{2}
\]

Would you like any further clarification or details on this solution?

Here are five related questions you might find useful:
1. What are the conditions for applying L'Hopital's Rule?
2. How do you differentiate trigonometric functions like \(\sin(x)\) and \(\cos(x)\)?
3. What other methods can be used to evaluate limits that result in indeterminate forms?
4. How can we simplify trigonometric expressions before applying limits?
5. What are some real-world applications of limits and indeterminate forms?

**Tip:** Always check for indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) before applying L'Hopital's Rule to avoid incorrect assumptions about the limit's value.

If g is the function defined by g(x) = (cos(x) - sin(x)) / (1 - 2sin^2(x)), what is lim(x→π/4) g(x)?

Discover how to solve the limit of g(x) = (cos(x) - sin(x)) / (1 - 2sin^2(x)) as x approaches π/4. This calculus problem involves applying L'Hopital's Rule to handle the indeterminate form 0/0, using trigonometric identities. Suitable for college-level calculus students.

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