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

Answer 1

#oo#

After factorizing the fraction's terms, we will simplify:

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

The limit will then become apparent to us:

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

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 the right:

Let #t= -0.9999#

Then:

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

Proceeding left and right will bring you closer and closer.

#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
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Answer 3

To find the limit of (t^2+t-2)/(t^2-1) as t approaches -1, we can substitute -1 into the expression and simplify.

Substituting -1 into the expression, we get (-1^2+(-1)-2)/((-1)^2-1).

Simplifying further, we have (1-1-2)/(1-1), which becomes -2/0.

Since division by zero is undefined, the limit does not exist.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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