### 5.11. Compute the limit

$$
\lim _{x \rightarrow 0} \frac{1+\sin 2 x-\cos 2 x}{1-\sin 2 x-\cos 2 x}
$$
Solution. Under the limit sign, we have an indeterminate form of type "0/0". To resolve this, we will use the known trigonometric formulas:

$$
\sin 2 x=2 \sin x \cos x, \quad 1-\cos 2 x=2 \sin ^{2} x
$$

Taking this into account, we get

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1+\sin 2 x-\cos 2 x}{1-\sin 2 x-\cos 2 x}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} x+2 \sin x \cos x}{2 \sin ^{2} x-2 \sin x \cos x}= \\
& =\lim _{x \rightarrow 0} \frac{2 \sin x(\sin x+\cos x)}{2 \sin x(\sin x-\cos x)}=\lim _{x \rightarrow 0} \frac{\sin x+\cos x}{\sin x-\cos x}=\frac{0+1}{0-1}=-1 .
\end{aligned}
$$