Limit with both cosine and sine

Question

Evaluate $$\lim_{x \to \frac{\pi}{4}}\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}$$ I tried substituting and then series expansion but I don't think it is working out.

Answer 1

$$\cos x+\sin x=\sqrt2\cos\left(x-\frac\pi4\right)=\sqrt2\cos\frac{4x-\pi}4$$

$$\sqrt2-(\cos x+\sin x)=\sqrt2\left[1-\cos\frac{4x-\pi}4\right]=2\sqrt2\sin^2\left(\frac{4x-\pi}8\right)$$ (using $\cos2A=1-2\sin^2A$)

Finally use, $\lim_{h\to0}\dfrac{\sin h}h=1$

Answer 2

Write $x-\dfrac\pi4=2y$ to get $$\lim_{y\to0}\frac{\sqrt2-\cos\left(2y+\dfrac\pi4\right)-\sin\left(2y+\dfrac\pi4\right)}{(8y)^2}$$

$$=\frac{\sqrt2}{64}\lim_{y\to0}\frac{1-\cos2y}{y^2}$$

$$=\frac{\sqrt2}{64}\lim_{y\to0}\frac{2\sin^2y}{y^2}=\cdots$$

Answer 3

Let $x=u+\pi/4$. We have $\cos(u+\pi/4)+\sin(u+\pi/4)=\sqrt2\cos(u)$. $$ \begin{align} &\lim_{x\to\pi/4}\frac{\sqrt2-\cos(x)-\sin(x)}{(4x-\pi)^2}\\ &=\lim_{u\to0}\frac{\sqrt2(1-\cos(u))}{16u^2}\frac{1+\cos(u)}{1+\cos(u)}\\ &=\lim_{u\to0}\frac{\sin^2(u)}{u^2}\frac{\sqrt2}{16}\frac1{1+\cos(u)}\\ \end{align} $$