Lim .... .... x-2
x->-2 ....x^2-4
I got 4 using algebra, but I did the thing where you put -1.999 instead of -2 into the function and got 1/2, and I tried an online limit calculator and got infinity...im so confused.
Apply Hospital's rule and you get $\displaystyle \frac{-1}{4}$:
$\displaystyle \lim_{x\rightarrow-2}\frac{x-2}{x^2 - 4} =
\lim_{x\rightarrow-2}\frac{\frac{d}{dx}(x-2)}{\frac{d}{dx}(x^2 - 4)} =
\lim_{x\rightarrow-2}\frac{1}{2x} =
\frac{1}{2(-2)} = \frac{-1}{4}
$
When you apply Hospital's rule you take the derivative of the top over the derivative of the bottom.
That notation: $\displaystyle \frac{d}{dx}$ just means that I'm taking the derivative of those two as if they were their own monomial, and ignoring the fact that one is actually in the denominator. Just a notation I use.
L'hospital's rule is.
$\displaystyle \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} $
if $\displaystyle \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0 $
then $\displaystyle \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)} $
Can L'hospital's rule be used here?
Actually, you can't, since the numerator doesn't approach either infinity or 0 when $\displaystyle x\to -2$
In fact, the limit $\displaystyle \lim_{x \to -2}\frac{x-2}{x^2-4}$ does not exist:
$\displaystyle \lim_{x \to -2}\frac{x-2}{x^2-4} = \lim_{x \to -2}\frac{1}{x+2}$
But,
$\displaystyle \lim_{x \to -2^+}\frac{1}{x+2} = +\infty$
And: $\displaystyle \lim_{x \to -2^-}\frac{1}{x+2} = -\infty$
Thus, the limit does not exist when $\displaystyle x\to -2$.
L'Hopital's rule is not applicable here as the numerator is not indeterminate at $\displaystyle x=-2$, the limit does not exist (it is not $\displaystyle +$ or $\displaystyle -$ infty as the sign as $\displaystyle x$ approaches $\displaystyle -2$ from the left and right is different)
CB