Kan418 Posts: 4, Reputation: 1 New Member #1 Jul 17, 2009, 03:32 AM
evaluate the following limits
lim x2(squared) - 6x + 8 / 2x2(squared) - 32
x-4 (the - is an arrow)
 iLLyriC Posts: 36, Reputation: 1 Junior Member #2 Jul 17, 2009, 05:52 AM
the easiest way is always L'ohspitals Rule:
Take the derivative of the numerator and denominator and substitute x, in this case:

lim (x^2 - 6x +8)/2x^2 - 32 = lim (2x -6)/4x = 2/16 = 1/2
x->4
 ebaines Posts: 12,132, Reputation: 1307 Expert #3 Jul 17, 2009, 08:47 AM
Originally Posted by iLLyriC
the easiest way is always L'ohspitals Rule:
Take the derivative of the numerator and denominator and substitute x, in this case:

lim (x^2 - 6x +8)/2x^2 - 32 = lim (2x -6)/4x = 2/16 = 1/2
x->4
Right, except 2/16 = 1/8, not 1/2.
 iLLyriC Posts: 36, Reputation: 1 Junior Member #4 Jul 17, 2009, 02:01 PM
Originally Posted by ebaines
Right, except 2/16 = 1/8, not 1/2.
LoL, thanks for correcting
 galactus Posts: 2,271, Reputation: 282 Ultra Member #5 Jul 17, 2009, 02:14 PM
$\lim_{x\to 4}\frac{x^{2}-6x+8}{2x^{2}-32}$

If I may, L'Hopital is not needed in this case. But it works OK.

We are only dealing with a rational function.

Notice we can factor:

$\lim_{x\to 4} \;\ \frac{(x-4)(x-2)}{2(x-4)(x+4)}=\lim_{x\to 4} \;\ \frac{(x-2)}{2(x+4)}=\frac{1}{8}$
 Kan418 Posts: 4, Reputation: 1 New Member #6 Jul 20, 2009, 03:08 AM
Thank you Galactus.

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