Limit with both cosine and sine 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.
 A: $$\cos x+\sin x=\sqrt2\cos\left(x-\frac\pi4\right)=\sqrt2\cos\frac{4x-\pi}4$$
$$\sqrt2-(\cos x+\sin x)=\sqrt2\left[1-\cos\frac{4x-\pi}4\right]=2\sqrt2\sin^2\left(\frac{4x-\pi}8\right)$$  (using $\cos2A=1-2\sin^2A$)
Finally use, $\lim_{h\to0}\dfrac{\sin h}h=1$
A: Write $x-\dfrac\pi4=2y$  to get $$\lim_{y\to0}\frac{\sqrt2-\cos\left(2y+\dfrac\pi4\right)-\sin\left(2y+\dfrac\pi4\right)}{(8y)^2}$$
$$=\frac{\sqrt2}{64}\lim_{y\to0}\frac{1-\cos2y}{y^2}$$
$$=\frac{\sqrt2}{64}\lim_{y\to0}\frac{2\sin^2y}{y^2}=\cdots$$
A: Let $x=u+\pi/4$. We have $\cos(u+\pi/4)+\sin(u+\pi/4)=\sqrt2\cos(u)$.
$$
\begin{align}
&\lim_{x\to\pi/4}\frac{\sqrt2-\cos(x)-\sin(x)}{(4x-\pi)^2}\\
&=\lim_{u\to0}\frac{\sqrt2(1-\cos(u))}{16u^2}\frac{1+\cos(u)}{1+\cos(u)}\\
&=\lim_{u\to0}\frac{\sin^2(u)}{u^2}\frac{\sqrt2}{16}\frac1{1+\cos(u)}\\
\end{align}
$$
A: This works just fine with a series expansion, but you have to use the expansion around $x = \pi/4$, not the usual series expansions around $x = 0$.  Working out the first few terms of these Taylor series,
$$ \eqalign{
   \sin x & = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \pi/4) - \frac{\sqrt{2}}{4} (x - \pi/4)^2 - \frac{\sqrt{2}}{12} (x - \pi/4)^3 + \frac{\sqrt{2}}{48} (x - \pi/4)^4 + \cdots \text,\\
   \text{and }\cos x & = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} (x - \pi/4) - \frac{\sqrt{2}}{4} (x - \pi/4)^2 + \frac{\sqrt{2}}{12} (x - \pi/4)^3 + \frac{\sqrt{2}}{48} (x - \pi/4)^4 - \cdots \text.}
   $$
Then the numerator becomes
$$ \sqrt{2} - \cos x - \sin x = \frac{\sqrt{2}}2 (x - \pi/4)^2 - \frac{\sqrt{2}}{24} (x - \pi/4)^4 + \cdots \text, $$
and the denominator becomes
$$ (4x - \pi)^2 = 16(x - \pi/4)^2 \text; $$
now you can drop insignificant terms, cancel factors, and get your answer.
I know that your tag said to avoid L'Hôpital's Rule, but if you notice that this rule works after 2 steps, then series expansion should also work with the most significant terms having degree 2.  Just be sure to expand around your limit point.
