Find the following limit problem I'm trying to find the following limit 
$$ \lim_{x\to 0} \left({1-\sin x } {\cos x}\right)^{{ \csc 2x} }$$
How to prove the above limit equals $e^{-{1\over2}}$?
 A: HINT:
$$(1-\sin(x)\cos(x))^{\csc(2x)}=\left(1-\frac1{2\csc(2x)}\right)^{\csc(2x)}$$
Let $t=2\csc(2x)$.  Then, we have
$$\lim_{x\to 0^+}(1-\sin(x)\cos(x))^{\csc(2x)}=\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^{\frac t2}$$
A: $$ \lim_{x\to 0} \left(1-\sin x \cos x\right)^{\csc 2x}=\lim_{x\to 0} \left(1-\sin x \cos x\right)^{\frac{1}{\sin 2x}}=$$
$$ \lim_{x\to 0} \left(1-\sin x \cos x\right)^{\frac{1}{\sin x\cos x}\frac{1}{2}}=e^{-1\cdot\frac{1}{2}}$$
A: HINT:
$(1-\sin x \cos x)^{\csc 2x}=(1-\sin x \cos x)^{\frac{1}{\sin 2x}}=\left[(1-\sin x \cos x)^{\frac{1}{\sin x \cos }}\right]^{\sin x \cos x \cdot \frac{1}{\sin 2x}}=\left[(1-\sin x \cos x)^{\frac{1}{\sin x \cos }}\right]^{\sin x \cos x \cdot \frac{1}{2 \sin x \cos x}}=\left[(1-\sin x \cos x)^{\frac{1}{\sin x \cos }}\right]^{\frac{1}{2 }}$
A: The way the formula appears,
$$
\begin{align}
\lim_{x\to 0}\left(1-\sin(x)\cos(x)\right)^{\csc(2x)}
&=\lim_{x\to 0}\left(1-\frac12\sin(2x)\right)^{\csc(2x)}\\
&=\lim_{x\to 0}\left(1-\frac1{2\csc(2x)}\right)^{\csc(2x)}\\[4pt]
&=e^{-1/2}
\end{align}
$$
However, the LaTeX seems to indicate that the problem may have meant to ask
$$
\begin{align}
&\lim_{x\to 0}\left((1-\sin(x))\cos(x)\right)^{\csc(2x)}\\
&=\lim_{x\to 0}\left(1-\sin(x)\right)^{\csc(2x)}\cos(x)^{\csc(2x)}\\
&=\lim_{x\to 0}\left(1-\sin(x)\right)^{\csc(2x)}\left(1-\sin^2(x)\right)^{\csc(2x)/2}\\
&=\lim_{x\to 0}\left(1-\sin(x)\right)^{3\csc(2x)/2}\left(1+\sin(x)\right)^{\csc(2x)/2}\\
&=\lim_{x\to 0}\left(1-\frac1{\csc(x)}\right)^{3\csc(x)\sec(x)/4}\lim_{x\to 0}\left(1+\frac1{\csc(x)}\right)^{\csc(x)\sec(x)/4}\\[4pt]
&=e^{-3/4}e^{1/4}\\[9pt]
&=e^{-1/2}
\end{align}
$$
Both interpretations give the desired answer since as $x\to0$, $\csc(x)\to\infty$.
