# Math Help - Limit Problem

1. ## Limit Problem

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

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

It equals -1/4

3. This is what it looks like right?

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

4. Hey sorry x approaches 1 instead of 2

5. Originally Posted by Sturm88
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 ......

6. 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