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.

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- Aug 30th 2009, 07:05 PMusmelikcheesFinding this limit using Algebra
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. - Aug 30th 2009, 07:34 PMpickslides
$\displaystyle \lim_{x \to 2}\frac{x-2}{x^2-4}$

$\displaystyle \lim_{x \to 2}\frac{x-2}{(x-2)(x+2)}$

$\displaystyle \lim_{x \to 2}\frac{1}{x+2} = \frac{1}{2+2}= \frac{1}{4}$ - Aug 30th 2009, 07:36 PMusmelikchees
- Aug 30th 2009, 08:22 PMeXist
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}

$ - Aug 30th 2009, 09:09 PMusmelikchees
- Aug 30th 2009, 09:14 PMeXist
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. - Aug 30th 2009, 09:28 PMusmelikchees
- Aug 30th 2009, 09:34 PMpickslides
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? - Aug 31st 2009, 01:11 PMusmelikchees
- Aug 31st 2009, 01:58 PMDefunkt
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 $\displaystyle x\to -2$. - Aug 31st 2009, 01:58 PMCaptainBlack
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