# Thread: Another question on Limits

1. ## Another question on Limits

Given:

$\lim_{\Delta x \to 2}$

$2-x$
___________________
$x^2-4$

What would I do to solve down on this?

2. Hello buddy.

Originally Posted by jimmyp
Given:

$\lim_{\Delta x \to 2}$

$2-x$
___________________
$x^2-4$

What would I do to solve down on this?
Hint: (x-2)(x+2) = x^2-4

= $lim \frac{2-x}{(x-2)(x+2)}$

Note that 2-x = - (x-2)

= $lim \ -\frac{x-2}{(x-2)(x+2)} = - lim \frac{1}{x+2} = -1/4$

x to +2

Cheers
Rapha

3. Sorry the lim Delta x -> 2 was suppose to read just lim x -> 2

I don't believe your technique still holds in that case?

4. Originally Posted by jimmyp
Sorry the lim Delta x -> 2 was suppose to read just lim x -> 2

I don't believe your technique still holds in that case?
Why would the factoring and cancellation not work...? What did you get, when you tried to use that method, that led you to this conclusion?

5. Originally Posted by jimmyp
Sorry the lim Delta x -> 2 was suppose to read just lim x -> 2

I don't believe your technique still holds in that case?
Actually, because there is no " $\Delta x$" in the function, Rapha did that already assuming that you mean $\lim_{x\rightarrow 2} \frac{2- x}{x^2- 4}$

6. Originally Posted by HallsofIvy
Actually, because there is no " $\Delta x$" in the function, Rapha did that already assuming that you mean $\lim_{x\rightarrow 2} \frac{2- x}{x^2- 4}$
Sorry - I now see where my logic was flawed. Thank you both.