Evaluate $\displaystyle\lim_{x\to1}\left(\frac{1}{x^2+x-2} -\frac x{x^3-1}\right)$
I will tell you frankly that I am trying my best and have used synthetic division and got $$\frac{1}{(x+2)(x-1)} -\frac x{ (x-1)(x^2+x+1)}$$ but from here on my every attempt seems doubtful whether I am doing it right? Can anyone help?
On
$\require{cancel}$
$$\begin{align} \frac{1}{x^2+x-2}-\frac{x}{x^3-1}& = \frac{(x^2+ x +1) - (x+2)x}{(x+2)(x-1)(x^2+x+1)}\\ \\ &= \frac{-x+1}{(x+2)(x-1)(x^2 + x + 1)} \\ \\ &= \frac{-(\cancel{x-1})}{(x+2)(\cancel{x-1})(x^2 + x + 1)}\\ \\ &=\;-\left(\frac 1{(x+2)(x^2+x+1)}\right)\end{align}$$
$$ $$
$$\lim_{x\to 1}\;-\left(\frac 1{(x+2)(x^2+x+1)}\right) = -\frac{1}{(3)(3)}=-\frac 19$$
As I said in a comment above, please post the original problem, including your synthetic division.
Find the least common multiple of $(x^2+x-2)$ and $(x^3-1)$ first. One easy way is to first factor the two expressions, and then multiply the unique factors together. Then, convert both fractions to have a denominator equivalent to this LCM. You should be able to cancel factors out in the end. Eventually, you will find that all (x-1) terms are cancelled, and simply plugging in 1 will yield the limit.