Evaluate `lim_(x to (pi)/(4)) (cos x - "sin" x)/(cos 2x)`

Answer

To evaluate the limit

\[ \lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\cos 2x}, \]

we can follow these steps:

### Step 1: Substitute the limit into the expression
First, we will substitute \( x = \frac{\pi}{4} \) into the expression:

\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}. \]

Thus,

\[ \cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0. \]

Now we need to evaluate \(\cos 2x\) at \(x = \frac{\pi}{4}\):

\[ \cos\left(2 \cdot \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0. \]

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Simplify the expression
To resolve the indeterminate form, we can use the double angle identity for cosine:

\[ \cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x). \]

Now we can rewrite the limit:

\[ \frac{\cos x - \sin x}{\cos 2x} = \frac{\cos x - \sin x}{(\cos x - \sin x)(\cos x + \sin x)}. \]

### Step 3: Cancel the common terms
Since \(\cos x - \sin x\) is common in the numerator and denominator (and is not equal to zero for \(x\) near \(\frac{\pi}{4}\)), we can cancel it out:

\[ \lim_{x \to \frac{\pi}{4}} \frac{1}{\cos x + \sin x}. \]

### Step 4: Substitute again
Now we substitute \(x = \frac{\pi}{4}\) into the simplified expression:

\[ \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}. \]

### Step 5: Final calculation
Thus, we have:

\[ \lim_{x \to \frac{\pi}{4}} \frac{1}{\cos x + \sin x} = \frac{1}{\sqrt{2}}. \]

### Conclusion
The final result is:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\cos 2x} = \frac{1}{\sqrt{2}}. \]

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