Q:

if two functions are not equal the how come limit of those two are equal?

Suppose we are having two functions $$f(x )=\frac{\sin x(4-x^2)}{4x-x^3}$$ and $$g(x)=\frac{\sin x}{x}$$ they are not equal everywhere(for all real numbers), but they both are equal only in ${R}-${$0,2,-2$} 
so while finding continuity writing the function $\frac{4-x^2}{4x-x^3}$ as $\frac{1}{x}$ is wrong.
but when we are evaluating the limit why do we take them equal? for example $$\lim_{x\to 0}{\frac{\sin x(4-x^2)}{4x-x^3}}$$ we write it as $$\lim_{x\to 0}{\frac{\sin x}{x}} $$ this means $$\frac{\sin x(4-x^2)}{4x-x^3}=\frac{\sin x}{x}$$ 
and this equality is totally wrong because as i have written above $f(x)\ne g(x)$, then while evaluating limit why do i take these two to be equal, this means we are changing our functions while evaluating the limit, i mean when we have to find $\lim_{x\to 0}f(x)$ we convert our function and find $\lim_{x\to 0}g(x)$.
And how our limit is correct if i changed my function.
I know limit means:

"How does a function behave when it is close to some number"

but when we evaluate the limit we are changing our function(from $f(x)$ to $g(x)$)  then how can i expect that my limit of $f(x)$ is correct because i changed it to limit of $g(x)$
if i have misunderstood the meaning of limit, please do explain it to me :-)

A:

$$f(x)=\frac{\sin x(4-x^2)}{4x-x^3}=\frac{\sin x(4-x^2)}{x(4-x^2)} = \frac{\sin x}{x}, |x|\ne2$$
The function above is equal to $g(x)=\frac{\sin x}{x}$ everywhere but at the discontinuities $x=\pm2$. We see that through algebraic simplification, they have a very similar form. The major difference is in the domain of the functions: the former is not defined at $x=0,\pm2$, the latter is not defined at $x=0$. 
When evaluating the limit, you've only shown interest in approaching $x\rightarrow 0$. Well, the two functions are identical in that locality, so they have the same behavior. You may ask then, what happens near another interesting point? What about $x\rightarrow \pm2$? Would my functions be different there? The answer in this case is yes, your functions would the different: we showed that earlier with the domains. However, their behavior in the limit is still the same for this example. Why? The discontinuity was removable!

