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# ${\displaystyle \lim_{x\to\frac{\pi}{4}^-} \frac{\sqrt{1-\sin2x}}{x-\frac\pi4} }$

$$\newcommand{\d}{\mathrm{d}}
\newcommand{\alb}{\begin{align*}}
\newcommand{\ale}{\end{align*}}
\newcommand{\dis}{\displaystyle}
\newcommand{\red}{\color{red}}
\newcommand{\brown}{\color{brown}}
\newcommand{\lrr}[1]{\left( #1 \right)}
\newcommand{\lrs}[1]{\left[ #1 \right]}
\newcommand{\overequal}[1]{\stackrel{ #1 }{=}}
\newcommand{\lx}[1]{\lim_{x\to{ #1 }}}
$$

$$\alb
\lx{\frac\pi4^-} \frac{\sqrt{1-\sin2x}}{x-\frac\pi4}
&= \lx{0^-}\frac{\sqrt{1-\sin(2x+\frac{\pi}{2})}}{x}\\
\ale
$$

Note that:

:::warning
$$\alb
&\cos x = \sin(x + \frac\pi2)\\
\implies& \cos2x = \sin(2x + \frac\pi2)
\ale
$$

$$\alb
&\cos(\alpha+\beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta\\
\implies& \cos 2x = \cos^2x-\sin^2x = 1-2\sin^2x\\
\implies& \left| \sin x \right| = \sqrt{ \frac{1-\cos 2x}{2} } \\
\implies& \sqrt{ 1-\sin\!\lrr{2x + \frac\pi2} } = \sqrt{2} \left| \sin x \right|\\
\ale
$$
:::

For $-\pi \leq x \leq 0$,

$$\left| \sin x \right| =- \sin x
$$
So we get
$$\alb
\lx{0^-}\frac{\sqrt{1-\sin(2x+\frac{\pi}{2})}}{x}
&= \lx{0^-}\frac{-\sqrt{2}\sin x}{x}\\
&= -\sqrt{2}
\ale
$$