Solve $\lim_{x\to \pi/4} \frac{\sin x - \cos x}{x-\pi/4}$ As the title suggests, we have to solve the limit:
$\lim_{x\to \frac\pi4} \frac{\sin x - \cos x}{x-\frac \pi4}$
I'm able to solve it by using L'Hospital's rule and got an answer $\sqrt2$ but the problem is that this rule is not allowed at school level.
So I tried another method:
$$\lim_{x\to \frac\pi4} \frac{\sin x - \cos x}{x-\frac \pi4}$$
$$\lim_{h\to 0} \frac{\sin(π/4+h) - \cos(π/4+h)}{h}$$
By using the identity of $\sin(a+b)$ and $\cos(a+b)$, we get:
$$\lim_{h\to 0} \frac{[\sin π/4+ \cos π/4][\cos h + \sin h]}{h}$$
If we here substitute $h=0$, we get $√2/0$. Can we solve it further?
Please help!
BTW sorry for the bad formatting.
 A: As suggested in the comments by $t=x-\frac \pi 4\to 0$ and $\sin x - \cos x = \sqrt 2 \sin \left(x-\frac \pi 4\right)$ we have
$$\lim_{x\to \pi/4} \frac{\sin x - \cos x}{x-\pi/4}=\sqrt 2\,\lim_{t\to 0} \frac{\sin t }{t}$$
A: You can use the fact that $\sin\pi/4=\cos\pi/4$ to write
$${\sin x-\cos x\over x-\pi/4}={\sin x-\sin\pi/4\over x-\pi/4}-{\cos x-\cos\pi/4\over x-\pi/4}$$
and now recognize the limit of each piece, separately, as the definition of the derivative $\sin x$ and $\cos x$, respectively, evaluated at $x=\pi/4$. Thus
$$\lim_{x\to\pi/4}{\sin x-\cos x\over x-\pi/4}=\lim_{x\to\pi/4}{\sin x-\sin\pi/4\over x-\pi/4}-\lim_{x\to\pi/4}{\cos x-\cos\pi/4\over x-\pi/4}=\cos\pi/4+\sin\pi/4=\sqrt2$$
(This is just a minor variant on José Carlos Santos's answer.)
A: Note that $\cos$ is continuous at $\pi/4$ and $\cos(\pi/4)>0$, so $\cos(x)$ is certainly nonzero near $\pi/4$. This justifies writing
\begin{align}
\frac{\sin(x)-\cos(x)}{x-\frac{\pi}{4}} &= \frac{\cos(x)\left(\frac{\sin(x)}{\cos(x)}-1\right)}{x-\frac{\pi}{4}}\\
&= \cos(x)\cdot\frac{\tan(x)-1}{x-\frac{\pi}{4}}\\
&= \cos(x)\cdot\frac{\tan(x)-\tan\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}
\end{align}
for every $x$ sufficiently close to $\pi/4$.
As $x\to\pi/4$, $\cos(x)\to\cos(\pi/4)=\sqrt{2}/2$ and
\begin{align}
\frac{\tan(x)-\tan\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}} &\to  (\tan)'\left(\frac{\pi}{4}\right)\\
&= \sec^2\left(\frac{\pi}{4}\right)\\
&= 2
\end{align}
It follows from the product rule for limits that
\begin{align}
\lim_{x\to\frac{\pi}{4}}\frac{\sin(x)-\cos(x)}{x-\frac{\pi}{4}} &= \lim_{x\to\frac{\pi}{4}}\left(\cos(x)\cdot\frac{\tan(x)-\tan\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}\right)\\
&= \frac{\sqrt{2}}{2}\cdot 2\\
&=\sqrt{2}
\end{align}
A: As an alternative by Euler's identity
$$\lim_{x\to \frac\pi4} \frac{\sin x - \cos x}{x-\frac \pi4}=\lim_{x\to \frac\pi4} \frac{e^{ix}-e^{-ix} -ie^{ix}+ie^{-ix}}{i\left(2x-\frac \pi2\right)}=$$
$$=\lim_{x\to \frac\pi4} \frac{1-i}{e^{ix}}\frac{e^{\left(2ix\right)}-i}{i\left(2x-\frac \pi2\right)}=\lim_{x\to \frac\pi4} \frac{1+i}{e^{ix}}\frac{e^{i\left(2ix-\frac \pi 2\right)}-1}{i\left(2x-\frac \pi2\right)}=\frac{1+i}{\frac{1+i}{\sqrt 2}} \cdot 1=\sqrt 2$$
