Clarify difference in solving limits to infinity vs finite limits I am looking to clarify the approach to solve the following limits:
f(x) = $\frac {4x - 2x^2}{x^2 - 4}$ find $\lim \limits_{x\to \infty}$ f(x)
and
$\lim \limits_{x\to 2}$ f(x)
To solve for $\lim \limits_{x\to \infty}$ f(x) I can divide by the largest power and arrive at:
$\lim \limits_{x\to \infty} \frac {4/x - 2}{1 - 4/x}$ = -2
Solving for $\lim \limits_{x\to 2}$ I can factor:
$\lim \limits_{x\to 2}$ f(x) => $\lim \limits_{x\to 2} \frac {2x(2-x)}{(x-2)(x+2)}$ => $\lim \limits_{x\to 2} \frac {-2x}{2-x}$ = -1
My question is: why can't I use the first method to solve for $\lim \limits_{x\to 2}$? It results in 0/0.
Thank you
 A: $\require{cancel}$
In the first case you could have factored first; it would not have been wrong to have done so:
$$\frac {4x - 2x^2}{x^2 - 4} = = \frac{2x(2-x)}{(x-2)(x+2)} = \frac{-2x(x-2)}{(x-2)(x+2)}$$
Alas, a common factor in the numerator and denominator.  Canceling that factor:
$$ = \frac{-2x(\cancel{x-2})}{(\cancel{x-2})(x+2)} = \frac{-2x}{x+2}$$
Now, taking $$\lim_{x\to \infty} \frac{-2x}{x+2},$$ we need to divide the numerator and denominator by $x$ to get:
$$\lim_{x\to \infty} \frac{-2}{1+\frac 2x} = -2.$$
Since factoring and simplifying first, or not doing so, both require the division of numerator and denominator by the largest power of $x$, it is likely more efficient to not bother with simplifying first.

In the second case, as $\lim_{x\to 2}$, factoring and simplifying the function gets immediate results:
$$\lim_{x\to 2} \frac{-2x}{x+2} = -\frac 44 = -1$$
just as you found.

In the second case, using the first approach will require much more work.  Let's see what happens:
$$\lim_{x\to 2} \frac {4x - 2x^2}{x^2 - 4}$$
Dividing this (unsimplified) function by the largest power of $x$, which is $x^2$ gives us $$\lim_{x\to 2}\frac{\frac 4x -  2}{1 - \frac 4{x^2}} = \frac {2-2}{1-\frac 44} \to \frac 00$$  which gets us nowhere.
The key about limits $x\to \infty$, e.g. $\lim_{x\to \infty}{\frac 4x} = 0$ annihilates the fraction, whereas $\lim_{x\to 2}{\frac 4x} = 2.$
So the best route for approaching the second limit, is to simplify as much as you can.  If factors cancel, all the easier to deal with.  If L'hopitals is needed, subsequently, after simplifying, it will be far easier to apply it to the simplified function, than to go back to step one.
A: use that $$\frac{4x-2x^2}{x^2-4}=\frac{2x(2-x)}{(x-2)(x+2)}$$
A: If you want to solve by that method you will have to use L'Hospital rule for indeterminate form since $\frac{0}{0} $is indeterminate which means we cannot tell it's value. I think you can solve it by seeing some notes or YouTube video. 
