<s>[INST] 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.
 [/INST] 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.
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