Limit of function. How can it suddenly change it's domain after simple manipulations I'm trying to refresh my math at the moment and have quickly become very confused by the calculation of limits of functions.
For example, I solved the following 
$\lim_{x \to 0} \frac{7x^2+4x^4}{3x^3-2x^2}$
by first manipulating it to 
$\frac{7+4x^2}{3x-2}$ and then concluding that the limit is $-\frac{7}{2}$
The thing I don't understand is why the original expression isn't defined in f(0) while the second one is? I'm not very experienced with math but I don't understand why the domain of the function can be changed by just multiplying the numerator and denominator of a fraction with the same value (which in this case is $\frac{1}{x^2}$).
 A: You are dividing with $x^2$ which lies in the denominator and hence $x\neq0$. So it can't be $f(0)$ anymore.
A: Yes, the domain did change, and for that reason they are not the same function. What you did here was cancel the factor $x^2$ off the numerator and denominator. Previously, because that factor of $x$ was on both the numerator and denominator we had what was called a removable discontinuity. These are points where a function is undefined but it looks as if a point has just been plucked off the graph and is denoted by an open circle. (Note if the $x$ was only on the denominator the function would have approached $\pm\infty$ and it would not be a removable discontinuity). So really
$$\frac{7x^2+4x^4}{3x^3-2x^2}\neq \frac{7+4x^2}{3x-2}$$
However you can say
$$\frac{7x^2+4x^4}{3x^3-2x^2}= \frac{7+4x^2}{3x-2}\qquad x\neq 0$$
But since we are dealing with a limit as $x\to 0$ we never have $x=0$ so this is not of concern here.
Also in response. Realize that you are multiplying the fraction by
$$\frac{x^{-2}}{x^{-2}}=1\qquad x\neq 0$$
But that multiplier is not defined at $x=0$, for that reason we must carry that information over into our simplified equation for the two new expressions to be truly equal.
