Show if $\lim_{x \rightarrow 0} \frac{\sin(\sin(x))}{x}$ exists, with product/quotient/substitution rules Show if $\lim_{x \rightarrow 0} \frac{\sin(\sin(x))}{x}$ exists, if so determine its value.
We have to use the product-rule and quotient-rule etc. to show this.
I thought this didn't exist, but my solution book says it does(without any explanation).
Here's why I thought it wouldn't exist:
Since $\lim_{x \rightarrow 0} \sin(x)=0$ and $\lim_{y \rightarrow 0} \sin(y)=0$ , using the substitution-rule gives us $\lim_{x \rightarrow 0} \sin(\sin(x))=0$.
We also know $\lim_{x \rightarrow 0} \frac{1}{x}$ diverges.
So when using the product rule, we find  $\lim_{x \rightarrow 0} \frac{\sin(\sin(x))}{x}$ doesn't exist.
I don't know where I went wrong here.
 A: Since the OP is not allowed to use L'Hôpital's rule, the following known limit can be used:
$$
\lim_{x\to 0} \frac{ \sin x}{  x}=1
$$
If we multiply numerator and denominator by $\sin x$ we obtain:
$$
\lim_{x\to 0} \frac{\sin (\sin x)}{x}=\lim_{x\to 0}\frac{\sin x \sin (\sin x)}{x \sin x}
$$
Now compute the two limits:
$$
\lim_{x\to 0} \frac{ \sin (\sin x)}{ \sin x}=\lim_{y\to 0} \frac{ \sin y}{ y}=1
$$
where we have called $y=\sin x$, and:
$$
\lim_{x\to 0} \frac{ \sin x}{  x}=1
$$
and apply the product rule, which is now allowed because both limits exist.
A: You are allowed to use the product rule when the limits of the factors exist, which is not the case here. Your use of the rule is invalid.
Actually, this is a $\dfrac00$ undeterminate form that you may not rewrite as $$\lim\frac00=\lim0\cdot\lim\frac10.$$

On the opposite, the solution given by Miguel Atencia is valid because in
$$\lim\frac{\sin(\sin(x))}{x}=\lim\frac{\sin(\sin(x))}{\sin(x)}\cdot\lim\frac{\sin(x)}{x}$$ both limits exist (and are $1$).
