میدانیم ${{(\sin x+\cos x)}^{2}}=1+\sin 2x$ زیرا:
${{(\sin x+\cos x)}^{2}}=\underbrace{{{\sin }^{2}}x+{{\cos }^{2}}x}_{1}\,\underbrace{2\sin x\,\cos x}_{\sin 2x}=1+\sin 2x$
از طرفی میدانیم $\sin x+\cos x=\sqrt{2}\sin \left( x+\frac{\pi }{4} \right)$ پس:
$\begin{align}
& 1|+\sin 2x={{(\sin x+\cos x)}^{2}}={{\left( \sqrt{2}\sin \left( x+\frac{\pi }{4} \right) \right)}^{2}}=2{{\sin }^{2}}\left( x+\frac{\pi }{4} \right) \\
& \underset{x\to \left( -\frac{\pi }{4} \right)}{\mathop{\lim }}\,\frac{1+\sin 2x}{{{\sin }^{2}}\left( x+\frac{\pi }{4} \right)}=\underset{x\to \left( -\frac{\pi }{4} \right)}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\left( x+\frac{\pi }{4} \right)}{{{\sin }^{2}}\left( x+\frac{\pi }{4} \right)} \\
\end{align}$
سعید احمدوند 
