# Finding this limit using Algebra

• Aug 30th 2009, 08:05 PM
usmelikchees
Finding 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, 08:34 PM
pickslides
$\lim_{x \to 2}\frac{x-2}{x^2-4}$

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

$\lim_{x \to 2}\frac{1}{x+2} = \frac{1}{2+2}= \frac{1}{4}$
• Aug 30th 2009, 08:36 PM
usmelikchees
Quote:

Originally Posted by pickslides
$\lim_{x \to 2}\frac{x-2}{x^2-4}$

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

$\lim_{x \to 2}\frac{1}{x+2} = \frac{1}{2+2}= \frac{1}{4}$

That would be right but the question is asking as x approaches -2 :p
• Aug 30th 2009, 09:22 PM
eXist
Apply Hospital's rule and you get $\frac{-1}{4}$:

$\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, 10:09 PM
usmelikchees
Quote:

Originally Posted by eXist
Apply Hospital's rule and you get $\frac{-1}{4}$:

$\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}
$

Ok thanks for the answer if its not too much trouble can i ask what the d/dx thing is?
• Aug 30th 2009, 10:14 PM
eXist
When you apply Hospital's rule you take the derivative of the top over the derivative of the bottom.

That notation: $\frac{d}{dx}$ 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, 10:28 PM
usmelikchees
Quote:

Originally Posted by eXist
When you apply Hospital's rule you take the derivative of the top over the derivative of the bottom.

That notation: $\frac{d}{dx}$ 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.

Ah yes :) thank you very very much.
• Aug 30th 2009, 10:34 PM
pickslides
L'hospital's rule is.

$\lim_{x\to a} f(x)=\frac{g(x)}{h(x)}$

if $\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0$

then $\lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)}$

Can L'hospital's rule be used here?
• Aug 31st 2009, 02:11 PM
usmelikchees
Quote:

Originally Posted by pickslides
L'hospital's rule is.

$\lim_{x\to a} f(x)=\frac{g(x)}{h(x)}$

if $\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0$

then $\lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)}$

Can L'hospital's rule be used here?

Yes it can
• Aug 31st 2009, 02:58 PM
Defunkt
Quote:

Originally Posted by usmelikchees
Yes it can

Actually, you can't, since the numerator doesn't approach either infinity or 0 when $x\to -2$

In fact, the limit $\lim_{x \to -2}\frac{x-2}{x^2-4}$ does not exist:

$\lim_{x \to -2}\frac{x-2}{x^2-4} = \lim_{x \to -2}\frac{1}{x+2}$

But,

$\lim_{x \to -2^+}\frac{1}{x+2} = +\infty$
And: $\lim_{x \to -2^-}\frac{1}{x+2} = -\infty$

Thus, the limit does not exist when $x\to -2$.
• Aug 31st 2009, 02:58 PM
CaptainBlack
Quote:

Originally Posted by eXist
Apply Hospital's rule and you get $\frac{-1}{4}$:

$\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}
$

L'Hopital's rule is not applicable here as the numerator is not indeterminate at $x=-2$, the limit does not exist (it is not $+$ or $-$ infty as the sign as $x$ approaches $-2$ from the left and right is different)

CB