Another question on Limits

**jimmyp** · May 21st 2009, 09:12 PM

Given:

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

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

What would I do to solve down on this?

**Rapha** · May 21st 2009, 09:16 PM

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

**jimmyp** · May 22nd 2009, 11:03 AM

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?

**stapel** · May 22nd 2009, 11:33 AM

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?

**HallsofIvy** · May 22nd 2009, 02:53 PM

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 $\lim_{x\rightarrow 2} \frac{2- x}{x^2- 4}$

**jimmyp** · May 23rd 2009, 09:13 PM

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.

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