How do you find the limit of (t^2+t-2)/(t^2-1) as t->-1?

2 Answers
Oct 8, 2017

oo

Explanation:

We will factorize the terms of the fraction and then simplify:

(t^2+t-2)/(t^2-1)=(cancel((t-1))(t+2))/(cancel((t-1))(t+1))

Then we will be able to find the limit:

lim_(t->-1)(t+2)/(t+1)=(-1+2)/(-1+1)=1/0=oo

Oct 9, 2017

Undefined.

Explanation:

(t^2+t-2)/(t^2-1)

Factor numerator: ((t+2)(t-1))/(t^2-1)

t^2-1=(t+1)(t-1) difference of two squares.

((t+2)(t-1))/((t+1)(t-1)) => (t+2)/(t+1)

As t->-1 from left.

Let t = -1.0001

Then

(t+2)/(t+1)= (-1.0001+2)/(-1.0001+1)= 0.9999/-0.0001=-9999

From right:

Let t= -0.9999

Then:

(t+2)/(t+1)= (-0.9999+2)/(-0.9999+1)= 1.0001/0.0001=10001

Moving closer and closer from left and right will give.

lim_(t->-1^-)(t+2)/(t+1)=-oo

lim_(t->-1^+)(t+2)/(t+1)=oo

So:

lim_(t->-1)(t+2)/(t+1)= undefined