Question

Evaluate the following limit: \[ \lim_{\theta \to -\frac{\pi}{4}} \frac{\cos\theta + \sin\theta}{\theta + \frac{\pi}{4}}. \]

• A. \( 0 \)
• B. \( 1 \)
• C. \( \sqrt{2} \)
• D. \( \frac{1}{\sqrt{2}} \)

Hint:

For limits involving trigonometric functions, use appropriate substitutions and trigonometric identities to simplify the expression before evaluating the limit.

Answer 1

The Correct Option is C

We are to evaluate the limit: \[ \lim_{\theta \to -\frac{\pi}{4}} \frac{\cos\theta + \sin\theta}{\theta + \frac{\pi}{4}}. \] Step 1: Let \( h = \theta + \frac{\pi}{4} \). As \( \theta \to -\frac{\pi}{4} \), \( h \to 0 \). The limit transforms to: \[ \lim_{h \to 0} \frac{\cos\left(-\frac{\pi}{4} + h\right) + \sin\left(-\frac{\pi}{4} + h\right)}{h}. \] Step 2: Apply trigonometric identities. Using \( \cos(-x) = \cos x \) and \( \sin(-x) = -\sin x \), we have: \[ \cos\left(-\frac{\pi}{4} + h\right) = \cos\left(\frac{\pi}{4} - h\right) = \cos\frac{\pi}{4}\cos h + \sin\frac{\pi}{4}\sin h, \] \[ \sin\left(-\frac{\pi}{4} + h\right) = -\sin\left(\frac{\pi}{4} - h\right) = -(\sin\frac{\pi}{4}\cos h - \cos\frac{\pi}{4}\sin h) = -\sin\frac{\pi}{4}\cos h + \cos\frac{\pi}{4}\sin h. \] Summing the numerator: \[ \cos\left(-\frac{\pi}{4} + h\right) + \sin\left(-\frac{\pi}{4} + h\right) = (\cos\frac{\pi}{4} - \sin\frac{\pi}{4})\cos h + (\sin\frac{\pi}{4} + \cos\frac{\pi}{4})\sin h. \] Since \( \cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \): \[ \cos\left(-\frac{\pi}{4} + h\right) + \sin\left(-\frac{\pi}{4} + h\right) = \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\cos h + \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)\sin h = \frac{2}{\sqrt{2}}\sin h = \sqrt{2}\sin h. \] Step 3: Compute the limit. Substitute the simplified numerator back into the limit expression: \[ \lim_{h \to 0} \frac{\sqrt{2}\sin h}{h}. \] We know the standard limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \). Therefore, the limit is: \[ \sqrt{2} \cdot 1 = \sqrt{2}. \] Final Answer: The value of the limit is: \[ \boxed{\sqrt{2}}. \]