1. ## limits

lim x→2g(x)
whereg(x) = ( x − 2 /x^ 2 − 4) if x 2
g(x)= 1 if x = 2
this confuses me any help would be nice ....

2. ## Re: limits

As it is written, your function is \displaystyle \begin{align*} x - \frac{2}{x^2} - 4 \end{align*}, but I suspect you meant \displaystyle \begin{align*} \frac{x - 2}{x^2 - 4} \end{align*}. Which one is it? If it's the latter, you MUST use brackets where they are needed!

3. ## Re: limits

yes it's the latter

4. ## Re: limits

hmmm the part that confuses me is I break it down to (x-2)/(x-2)(x+2) => 1/(x+2)

then sub in x = 2 to become 1/4

what confuses me is the part if x ≠ 2 and g(x)= 1 if x = 2

I'm having trouble with that concept

5. ## Re: limits

Originally Posted by bee77
hmmm the part that confuses me is I break it down to (x-2)/(x-2)(x+2) => 1/(x+2)
then sub in x = 2 to become 1/4
what confuses me is the part if x ≠ 2 and g(x)= 1 if x = 2
I'm having trouble with that concept
NEVER, NEVER, NEVER Say that. It will earn you a failing mark.

The notation $x\approx a$ means that $x$ is close to $a$.
So $\displaystyle \lim_{x\to a}f(x)=L$ means that if $x\approx a$ but $x\ne a$ then $f(x)\approx L$.

Now $\dfrac{x-2}{x^2-4}=\dfrac{1}{x+2}$ unless $x=2$.

So $\displaystyle \lim_{x\to 2}\dfrac{x-2}{x^2-4}=\displaystyle \lim_{x\to 2}\dfrac{1}{x+2}=\dfrac{1}{4}$ because by definition of limit $x\ne 2$ so $\dfrac{x-2}{x^2-4}=\dfrac{1}{x+2}$

6. ## Re: limits

$\dfrac{x-2}{x^2-4}$ is not defined at $x=2$ because you are dividing by zero (which is not a valid operation). However, they defined $g(x) = \begin{cases}\dfrac{x-2}{x^2-4} & x\neq 2 \\ 1 & x=2\end{cases}$ to mean that if $x\neq 2$, then $g(x) = \dfrac{x-2}{x^2-4}$ and $g(2) = 1$.

A limit of a function has nothing to do with the value of that function at that point. This is what is known as a removable discontinuity.

7. ## Re: limits

Thanks Plato ,I'll remember that sub part for my upcoming exam ...and the explanation of the concepts with the close to symbol really helps ,
cheers!!

8. ## Re: limits

Thanks SlipEternal ,so the g(2)=1 would simply imply a different part of the graph ...to the x-2/x^2-4 = 1/4 limit...?

9. ## Re: limits

I don't know what you mean by "imply a different part of the graph". The graph of y= g(x) is the graph of 1/(x+ 2) except that the point (1, 1/4) is removed and the point (1, 1) is added. But you take x closer and closer to 1, but not equal to 1, (x, y) on the graph gets closer and closer to (1, 1/4).

10. ## Re: limits

Originally Posted by Prove It
[bee77], you MUST use brackets where they are needed!
Originally Posted by bee77
hmmm the part that confuses me is I break it down to (x-2)/(x-2)(x+2) => 1/(x+2)
(x - 2)/[(x - 2)(x + 2)] => 1/(x + 2), $\displaystyle x \ne 2$

Originally Posted by bee77
x-2/x^2-4 = ...
(x - 2)/(x^2 - 4) =

bee77, do you see how the brackets are needed?

11. ## Re: limits

$g(x)=\left\{\begin{matrix} \dfrac{x-2}{x^2-4}\, & \, x\ne2\\ 1 \, & \, x=2 \end{matrix}\right.$

graph of $g(x)$ attached ... can you now visualize why $\displaystyle \lim_{x \to 2} g(x) = \dfrac{1}{4}$?

12. ## Re: limits

That makes it better in my head on the graph skeeter ,
Thanks !

13. ## Re: limits

Thanks HallsofIvy , I was just overthinking it in my head , thanks for the help