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

Oct 8, 2017

$\infty$

#### Explanation:

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

$\frac{{t}^{2} + t - 2}{{t}^{2} - 1} = \frac{\cancel{\left(t - 1\right)} \left(t + 2\right)}{\cancel{\left(t - 1\right)} \left(t + 1\right)}$

Then we will be able to find the limit:

${\lim}_{t \to - 1} \frac{t + 2}{t + 1} = \frac{- 1 + 2}{- 1 + 1} = \frac{1}{0} = \infty$

Oct 9, 2017

Undefined.

#### Explanation:

$\frac{{t}^{2} + t - 2}{{t}^{2} - 1}$

Factor numerator: $\frac{\left(t + 2\right) \left(t - 1\right)}{{t}^{2} - 1}$

${t}^{2} - 1 = \left(t + 1\right) \left(t - 1\right)$ difference of two squares.

$\frac{\left(t + 2\right) \left(t - 1\right)}{\left(t + 1\right) \left(t - 1\right)} \implies \frac{t + 2}{t + 1}$

As $t \to - 1$ from left.

Let $t = - 1.0001$

Then

$\frac{t + 2}{t + 1} = \frac{- 1.0001 + 2}{- 1.0001 + 1} = \frac{0.9999}{-} 0.0001 = - 9999$

From right:

Let $t = - 0.9999$

Then:

$\frac{t + 2}{t + 1} = \frac{- 0.9999 + 2}{- 0.9999 + 1} = \frac{1.0001}{0.0001} = 10001$

Moving closer and closer from left and right will give.

${\lim}_{t \to - {1}^{-}} \frac{t + 2}{t + 1} = - \infty$

${\lim}_{t \to - {1}^{+}} \frac{t + 2}{t + 1} = \infty$

So:

${\lim}_{t \to - 1} \frac{t + 2}{t + 1} =$ undefined