Evaluate $\lim_{t \rightarrow \frac{\pi}{4}}(\frac{|\sin^{\alpha}(t)-\cos^{\alpha}(t)|}{|\sin(t)-\cos(t)|^{\alpha}}), \alpha \in [0,1]$. I came across the following limit in an exercise and was wondering how to solve it:
$\lim_{t \rightarrow \frac{\pi}{4}}(\frac{|\sin^{\alpha}(t)-\cos^{\alpha}(t)|}{|\sin(t)-\cos(t)|^{\alpha}}), \alpha \in (0,1]$
Could it have something to do with the modulus of inequality?
I tried all the usual methods (l'hopital and some trig identities) but can't seem to get the desired result (=0). Would be nice if someone could give me some tips :)
PS: For anyone wondering what the original question was, it was to show that: for the function $f(x) = |x|^\alpha$ on the interval $[0,1]$, with $\alpha \in (0,1]$. Show that $\exists C > 0$ s.t. the modulus of continuity of $f(x)$ satisfies:
$\omega(f,h)\leq Ch^\alpha$, $h > 0$.
Determine the minimal possible value of $C$. (The limit came from my attempt to show $C \geq 1 \Rightarrow \min\{C\} = 1$)
 A: You can try to compute the limit
$$
\lim_{t \rightarrow \frac{\pi}{4}^+}(\frac{|\sin^{\alpha}(t)-\cos^{\alpha}(t)|}{(\sin(t)-\cos(t))^{\alpha}}), \alpha \in (0,1].
$$
In this way you can get rid of the absolute value and just compute
$$
\lim_{t \rightarrow \frac{\pi}{4}^+}\frac{\sin^{\alpha}(t)-\cos^{\alpha}(t)}{(\sin(t)-\cos(t))^{\alpha}}, \alpha \in (0,1].
$$
I will compute this limit, and the same will apply if you compute the left limit. Applying L'Hopital rule:
$$
\lim_{t \rightarrow \frac{\pi}{4}^+}\frac{\sin^{\alpha}(t)-\cos^{\alpha}(t)}{\sin(t)-\cos(t)^{\alpha}}=
\lim_{t \rightarrow \frac{\pi}{4}^+}\frac{\alpha \sin^{\alpha-1}(t)\cos (t)+\alpha\cos^{\alpha-1}(t)\sin(t)}{\alpha(\sin(t)-\cos(t))^{\alpha-1} (\cos(t)+\sin(t))}.
$$
Note that since $\alpha \in (0,1]$, $\alpha -1<0$, and therefore
$$
\lim_{t \rightarrow \frac{\pi}{4}^+}\frac{\alpha \sin^{\alpha-1}(t)\cos (t)+\alpha\cos^{\alpha-1}(t)\sin(t)}{\alpha(\sin(t)-\cos(t))^{\alpha-1} (\cos(t)+\sin(t))}=\lim_{t \rightarrow \frac{\pi}{4}^+}(\sin(t)-\cos(t))^{1-\alpha}\frac{\alpha \sin^{\alpha-1}(t)\cos (t)+\alpha\cos^{\alpha-1}(t)\sin(t)}{\alpha (\cos(t)+\sin(t))}=0.
$$
A: Note that $|x|^\alpha$ is differentiable at $x=\frac1{\sqrt 2}$, so in particular Lipschitz. Thus,
$$ |x|^\alpha = \frac1{2^{\alpha/2}} + O\Big(x-\frac1{\sqrt 2}\Big), \quad \text{as }x\to \frac1{\sqrt 2}.$$
Since
\begin{align}\cos t &= \frac1{\sqrt 2} - \frac{t-\frac\pi4}{\sqrt 2} + O\Big(\left|t-\frac\pi4\right|^2\Big), \quad \text{and} \\ \sin t &= \frac1{\sqrt 2} + \frac{t-\frac\pi4}{\sqrt 2} + O\Big(\left|t-\frac\pi4\right|^2\Big), \quad \text{as }t\to \frac\pi{ 4}, \end{align}
the numerator is $O(t-\frac\pi4)$, while the denominator is eventually (as $t\to \frac\pi4$) bounded from below by $\frac12|t-\frac\pi4|^\alpha$. So the limit  is zero.
For the original problem: let $C_*$ be the infimum of all possible $C$s. If $\omega(f,h)\le C|h|^\alpha$, then since $|h|^\alpha = f(h)-f(0) \le \omega(f,h),$ we must have $C\ge 1$. On the other hand,  a standard argument shows $|f(x)-f(y)| \le |x-y|^\alpha$, so $C_*\le 1$. Hence, $C_*=1$.
