Thread: Compute the Limit: Question 1

1. Compute the Limit: Question 1

Is my answer to the following question correct?

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

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

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

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

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

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

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

$\displaystyle = 1$

2. Re: Compute the Limit: Question 1

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

3. Re: Compute the Limit: Question 1

Is this correct now?

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

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

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

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

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

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

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

$\displaystyle = \frac{3}{2}$

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

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