Evaluate $\underset{θ→π/4}{\lim}\frac{\sqrt{2}-\cos θ-\sin θ}{(4θ-π)^2}$.

$\frac{1}{16\sqrt{2}}$

$\underset{θ→\frac{π}{4}}{\lim}\frac{\sqrt{2}-\cos θ-\sin θ}{(4θ-π)^2}=\frac{\sqrt{2}-\sqrt{2}\cos(θ-π/4)}{16(θ-π/4)^2}$

Let $θ-\frac{π}{4}=t⇒\underset{t→0}{\lim}\frac{\sqrt{2}-\sqrt{2}\cos t}{16t^2}=\underset{t→0}{\lim}\frac{\sqrt{2}}{16}\frac{2\sin^2t/2}{t^2}=\frac{\sqrt{2}}{32}\underset{t→0}{\lim}\frac{\sin^2t/2}{t^2/4}=\frac{1}{16\sqrt{2}}$