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

Question

7373 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-08T21:10:21

#oo#

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#

Answer 2 · 2017-10-09T12:41:35

Undefined.

#(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