Use Algebraic Methods to determine the limit. \[ \lim _{\theta \rightarrow \pi / 4} \frac{\sin ^{2} \theta-\cos ^{2} \theta}{\sin \theta-\cos \theta} \]

To find the limit \[ \lim _{\theta \rightarrow \pi / 4} \frac{\sin ^{2} \theta-\cos ^{2} \theta}{\sin \theta-\cos \theta}, \] we start by substituting \(\theta = \frac{\pi}{4}\): - At \(\theta = \frac{\pi}{4}\), \(\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\). Thus, both the numerator and denominator equal zero: \[ \sin^2 \frac{\pi}{4} - \cos^2 \frac{\pi}{4} = \frac{1}{2} - \frac{1}{2} = 0 \] and \[ \sin \frac{\pi}{4} - \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0. \] Since we have an indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's rule, which says to take the derivative of the numerator and the derivative of the denominator. 1. The derivative of the numerator, \(\sin^2 \theta - \cos^2 \theta\), is: \[ \frac{d}{d\theta}(\sin^2 \theta - \cos^2 \theta) = 2\sin \theta \cos \theta + 2\cos \theta \sin \theta = 4\sin \theta \cos \theta. \] 2. The derivative of the denominator, \(\sin \theta - \cos \theta\), is: \[ \frac{d}{d\theta}(\sin \theta - \cos \theta) = \cos \theta + \sin \theta. \] Now, we apply L'Hôpital's rule: \[ \lim _{\theta \rightarrow \pi / 4} \frac{\sin ^{2} \theta-\cos ^{2} \theta}{\sin \theta-\cos \theta} = \lim _{\theta \rightarrow \frac{\pi}{4}} \frac{4\sin \theta \cos \theta}{\cos \theta + \sin \theta}. \] Substituting \(\theta = \frac{\pi}{4}\): - The numerator becomes \[ 4\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = 4 \cdot \frac{1}{2} = 2. \] - The denominator becomes \[ \cos \frac{\pi}{4} + \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}. \] Thus, we have: \[ \lim _{\theta \rightarrow \pi / 4} \frac{4\sin \theta \cos \theta}{\cos \theta + \sin \theta} = \frac{2}{\sqrt{2}} = \sqrt{2}. \] Finally, the limit is \[ \boxed{\sqrt{2}}. \]