1

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 $\frac{1}{x^2}$).

Lasse
  • 13

2 Answers2

0

You are dividing with $x^2$ which lies in the denominator and hence $x\neq0$. So it can't be $f(0)$ anymore.

Papa Delta
  • 458
0

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.

Will Fisher
  • 5,112