What is the error in this approach to evaluating the limit? $\lim_{x\to0}(\sin\sin x- x\sqrt[3]{1-x^2})/x^5$ I am having trouble with a small thing in a limit. I encountered the limit
$$\lim_{x\to0}\frac{\sin\sin x- x\sqrt[3]{1-x^2}}{x^5}$$
which can be solved rather easily by using Taylor expansion and calculating the quotient. The problem, though, arised when I tried the more natural approach - I tried to simplify the expression. When doing that, I simplified the fraction by separating it into the two terms, first of which is $\frac{\sin\sin x}{x^5}$. Then, I multiplied this term by $\frac{\sin x}{\sin x}$ which creates $\frac{\sin\sin x}{\sin x}×\frac{\sin x}{x^5}$. When trying to find the original limit with this expression, it is still the same as the original limit should be. Although, I noticed that when I continue by firstly taking $\lim_{x\to0}\frac{\sin\sin x}{\sin x}=1$, the overall limit then simplifies to
$$\lim_{x\to0}\frac{\sin x}{x^5}-\frac{\sqrt[3]{1-x^2}}{x^4}$$
This limit, though, is equal to $\infty$, which is contrary to the original result, which should be $\frac{19}{90}$ (by using Taylor). Can anyone tell me what creates this error? I was thinking maybe the multiplying of $\frac{\sin x}{\sin x}$, but since that is just $1$ (no limits taken there), it should not pose a problem. The problem just seems to arise when I use the rule $\lim f(x)g(x)= \lim f(x)\lim g(x)$ on the first term there. Is this rule invalid in this case? If so, why? What am I missing? Sorry if this is really trivial, I am pretty tired by now and am probably missing something stupidly simple, but I really don't see it.
In advance, thanks for your time.
 A: The problem is that you implicitly also applied the sum-of-limits rule:
$$
\lim_{x\to0}\frac{\sin\sin x- x\sqrt[3]{1-x^2}}{x^5} = \lim_{x \to 0} \left( \frac{\sin\sin x}{x^5} \right) - \lim_{x \to 0} \left( \frac{x \sqrt[3]{1 - x^2}}{x^5} \right).
$$
However, this rule does not work because the two limits do not exist as real numbers. (Or if you prefer: $\infty - \infty$ is an indeterminate form.)
It's necessary to apply this rule first before doing your trick of multiplying by $\frac{\sin x}{\sin x}$. Otherwise, you can't "distribute" the $\frac{\sin x}{\sin x}$ into the first term only -- you would have to multiply the second term by $\frac{\sin x}{\sin x}$ as well.
For a simpler example of what went wrong, consider
$$
\lim_{x \to \infty} (x - x) = 0.
$$
We can write this as
$$
\lim_{x \to \infty} \left( \frac{x+1}{x+1} \cdot x - x\right)
= \lim_{x \to \infty} \left( \frac{x}{x+1} \cdot (x+1) - x\right),
$$
then according to the logic you used, since $\frac{x}{x+1} \to 1$, we can simplify this to
$$
\lim_{x \to \infty} \left( (x + 1) - x\right) = 1.
$$
Essentially, the product-of-limit rules does not apply if you are only applying it to some subterm of the limit. For a limit of the form $\lim (A + B)$, you can't apply the product-of-limit rule to just $A$ on its own.
