Finding $\lim\limits_{x \to 0}\ \frac{\sin(\cos(x))}{\sec(x)}$

The problem is to find:
$\lim\limits_{x \to 0}\ \dfrac{\sin(\cos(x))}{\sec(x)}$

I rewrite the equation as follows:
$\lim\limits_{x \to 0}\ \dfrac{\sin(\cos(x))}{\dfrac{1}{\cos(x)}}$

And multiply by $\dfrac{\cos(x)}{\cos(x)}$, producing:
$\lim\limits_{x \to 0}\ \dfrac{\cos(x)*\sin(\cos(x))}{\dfrac{\cos(x)}{\cos(x)}}$

And rewrite as:
$\lim\limits_{x \to 0}\ \cos^2(x)\ \dfrac{\sin(\cos(x))}{\cos(x)}$

Which then becomes:
$\lim\limits_{x \to 0}\ \cos^2(x) * 1$

Which becomes 1. However, the answer is apparently $\sin(1)$. What am I doing wrong?

Edit: I found a different way to solve this, but I'm still not sure what I did wrong originally.

Solution 1:

As $\displaystyle x \to 0$, $\displaystyle \cos x \to 1$.

So you cannot use the limit $\displaystyle \lim_{h \to 0} \frac{\sin h}{h} = 1$.

The given answer is $\displaystyle \sin(1)$ I presume and not $\displaystyle \sin(0)$ (which is $0$)...

$\lim_{x \to 0}\ \cos^2(x)\ \dfrac{\sin(\cos(x))}{\cos(x)}=\lim_{x \to 0}\ \cos^2(x)\ \dfrac{\sin(1)}{1}=\lim_{x \to 0}\ \cos^2(x)\sin(1)=\sin(1)$

Simple mistake when simplifying the term $\dfrac{\sin(\cos(x))}{\cos(x)}$