To evaluate `lim_(x->1)(x^2+x-2)/(x^2-x)` we cannot set `x=1` because it makes the denominator zero.

We test the numerator to see if it become zero when `x=1` and it is. so it has a factor of `(x-1)` in common with denominator.

`lim_(x->1)(x^2+x-2)/(x^2-x)` = `((x-1)(x+2))/(x(x-1))`=`(x+2)/(x)` if `x!=1`.

now let's evaluate `x->` we get `lim_(x->1)(x^2+x-2)/(x^2-x)` = `lim_(x->1)(x+2)/x`= `3`

How to:

to have`lim_(x->1)` use \`lim_(x->1)\`

to have numerator and denominator show up correctly, it is strongly suggested to us ( and ) to group them the same way using your calculator.