When I can't convert a piece of a limit in a notable limit? When I solve limits I often convert a block in a notable limit multiplying and dividing it by the same quantity. For example:
$$
\lim_{x \to 0} \frac{e^{\sin(x)}-1}{x} = \lim_{x \to 0} \frac{e^{\frac{\sin(x)}{x}x}-1}{x} = \lim_{x \to 0} \frac{e^{x}-1}{x} = 1
$$
I've noticed that I can't always apply this. For example:
$$
\lim_{x \to 0} \frac{x^3+9x-9 \tan(x)}{-8x^3} = \frac{1}{4}\neq \lim_{x \to 0} \frac{x^3+9x-9 \frac{\tan(x)}{x}x}{-8x^3} = \lim_{x \to 0} \frac{x^3+9x-9x}{-8x^3}= -\frac{1}{8}
$$
What's wrong in that passage? What rule am I violating?
Thanks
 A: Intuitively, the problem is that $\tan x$ has a term in the Taylor series proportional to $x^3$ which you are ignoring, but the terms of interest are of order $x^3$.  You can ignore the terms of order $x^5$ and up because they will contribute $0$ in the limit.
A: I'll try expand on Ross' answer.
Functions like $\sin x$, $\tan x$ and $\exp x$ can be approximated by Taylor polynomials near $x=0$ (and usually around any $x=a$ that makes sense - note the delicate case for $\tan x$). 
What can be enlightening to consider in the subtitutions you are making is the order of the approximation relative to the greatest power of $x$. We say a function $\alpha(x)$ is an infinitesimal when $x \to a$ if 
$$\lim_{x \to a} \alpha(x) = 0$$
We say two infinitesimals are of the same order if
$$\lim \frac \alpha \beta =A\neq 0$$
If it is the case $A=1$, we say $\alpha$ and $\beta$ are equivalent infinitesimals and we write
$$\alpha \sim \beta$$
Given two infinitesimals $\alpha$ and $\beta$ in $x \to a$, we say $\beta$ is of greater order than $\alpha$ if 
$$\lim \frac \alpha \beta =0$$
This means, $\alpha$ is "smaller" at $x =a$ than $\beta$. 
Likewise, we say that $\beta$ is an infinitesimal of order $n$ respect to $\alpha$ if $\alpha^n$ and $\beta$ are of the same order.
$$\lim \frac \beta {\alpha^n} =A$$
As you note, there are some notable limits, such as
$$e^x -1\sim x $$
$$\sin x \sim x $$
$$\log (x+1) \sim x $$
These can all be proven from the Taylor series definition's or some other arifice. But the Taylor series tells us more. In fact, Taylor series tells us about the degree of approximation we obtain when we approximate a function, such as $\sin x$, by a certain $n$ degree polynomial. And the result is that 
If $T_n^a(f,x)$ is the Taylor polynomial of $f$, of $n$th degree around $x=a$, then 
$$f(x) -T_n^a(f,x)=o[(x-a)^n] \text{ ; } x \to a$$
Read: $f(x) -T_n^a(f,x)$ is an infinitesimal of smaller order than $(x-a)^n$ when $x \to a$, or $f(x) -T_n^a(f,x)$ is small-$o$ of $(x-a)^n$ when $x \to a$.
The idea, as you see, is that
$$\lim \frac{f(x) -T_n^a(f,x)}{(x-a)^n} =0 $$
The importance of this is that it tells us the error made by approximating $f$ by it's $n$th degree polynomial, is small in comparison to $(x-a)^n$. When we swap between functinons and their approximation when calculating limits, we have to make sure the approximation is of order smaller than the greatest degree of $x$ we see, otherwise, the error will not go to zero when we take the limit! In your example, you are implicitly noting that
$$\tan x+o(x) = x $$ 
You state that
$$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9\tan (x)}}{{ - 8{x^3}}} = \frac{1}{4}$$
$$\eqalign{
  & \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9\left( {x + o\left( x \right)} \right)}}{{ - 8{x^3}}} =   \cr 
  &  = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9x - 9o\left( x \right)}}{{ - 8{x^3}}} =   \cr 
  &  =  - \frac{1}{8} + \mathop {\lim }\limits_{x \to 0} \frac{9}{8}\frac{{o\left( x \right)}}{{{x^3}}} \cr} $$
And there is an error that doesn't go away, because it would only go away if the denomitor had been $x$, or a greater infinitesimal (1 for example). But is is $x^3$. To fix this, we use
$$\tan x = x + \frac{{{x^3}}}{3} + o\left( {{x^3}} \right)$$
This gives
$$\eqalign{
  & \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9\left( {x + \frac{{{x^3}}}{3} + o\left( {{x^3}} \right)} \right)}}{{ - 8{x^3}}}  \cr 
  & \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9x - 3{x^3} - 9o\left( {{x^3}} \right)}}{{ - 8{x^3}}}  \cr 
  & \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{x^3}}}{{ - 8{x^3}}} + \mathop {\lim }\limits_{x \to 0} \frac{9}{8}\frac{{o\left( {{x^3}} \right)}}{{{x^3}}} = \frac{1}{4} \cr} $$
which is correct.
A: The $x$'s always have to go "together" to the limit, or else you could do silly things like this: 
$\lim\limits_{x\rightarrow 0^+}\frac{x}{x}=?!=\lim\limits_{x\rightarrow 0^+}\frac{0}{x}=0$ when the real limit is actually 1.
