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#1

$Old$ 08-28-2008, 10:30 PM

Sturm88 $Sturm88 is offline$

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Lim t^3+t^2-5t-3/t^3-3t+2
x->1

Please help, can't seem to factor the top portion at all.

Last edited by Sturm88; 08-28-2008 at 11:35 PM.

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#2

$Old$ 08-28-2008, 10:49 PM

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You don't need too, just plug in 2 for (t)

It equals -1/4

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#3

$Old$ 08-28-2008, 10:51 PM

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This is what it looks like right?

$\frac{t^3+t^2-5t-3}{t^3-3t+2}$

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$Old$ 08-28-2008, 11:35 PM

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Hey sorry x approaches 1 instead of 2

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$Old$ 08-28-2008, 11:48 PM

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Quote:

Originally Posted by Sturm88 $View Post$

Hey sorry x approaches 1 instead of 2

Then the limit is equal to -oo.
Note that the denominator is $(x-1)^2(x+2)$ which is always positive for x > -2 and note that numerator is always negative for x > 0 ......

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$Old$ 08-29-2008, 04:22 AM

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hi Sturm88
conventional method for finding limit can also be used.(though its lengthy but simple and general)
put t=1+h and find limit at h->0+ this will give you right hand limit
put t=1-h and find limit at h->0+ this will give you left hand limit
when you will calculate them you will get limit=- infinity
that is
left hand limit=right hand limit=-infinity

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