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

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

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

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

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

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

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.

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I don't see how you got $\lim_{x\rightarrow0}\cos^{2}(x)\cdot1$ from the limit above it. You should also note that direct substitution gives you $\sin(1)$ – WWright Oct 15 '10 at 4:05
I think you need to look again at your very last step. – jericson Oct 15 '10 at 4:06
BTW I wish more users of this site put this much effort into asking their questions. Nice job – jericson Oct 15 '10 at 4:14
In response to the edit, jericson's first comment is exactly the point. – BBischof Oct 26 '10 at 6:07

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 $0$)...

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

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

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Taking limit in limit can be wrong generally. – vudu vucu Mar 13 at 21:22