# Thread: Compute the Limit: Question 1

1. ## Compute the Limit: Question 1

Is my answer to the following question correct?

$\text{Calculate }\lim_{x \rightarrow 1 } \frac{x^2+x-2}{x^2-1}$

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

$= \frac{0}{0 }\text{Indeterminate}$

$= \lim_{x \rightarrow 1 } \frac{x^2+x-2}{x^2-1}$

$= \lim_{x \rightarrow 1 } \frac{x-2}{-1}$

$= \frac{1-2}{-1}$

$= \frac{-1}{-1}$

$= 1$

2. ## Re: Compute the Limit: Question 1

$\displaystyle \lim_{x\to 1}\frac{x^2+x-2}{x^2-1} = \lim_{x\to 1}\frac{(x+2)(x-1)}{(x-1)(x+1)} = \lim_{x\to 1}\frac{x+2}{x+1}= \dots$

3. ## Re: Compute the Limit: Question 1

Is this correct now?

$\text{Calculate }\lim_{x \rightarrow 1 } \frac{x^2+x-2}{x^2-1}$

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

$= \frac{0}{0 }\text{Indeterminate}$

$= \lim_{x \rightarrow 1 } \frac{x^2+x-2}{x^2-1}$

$= \lim_{x \rightarrow 1 } \frac{(x+2) \cdot (x-1)}{(x+1) \cdot (x-1)}$

$= \lim_{x \rightarrow 1 } \frac{x+2}{x+1}$

$= \frac{1+2}{1+1}$

$= \frac{3}{2}$

$= 1\frac{1}{2}$

4. ## Re: Compute the Limit: Question 1

That is much better.

5. ## Re: Compute the Limit: Question 1

Originally Posted by pickslides
That is much better.
Thanks so much pickslides