Question

Evaluate $\underset{x\to 1}{lim}(\frac{1}{x^2+x-2}-\frac{x}{x^3-1})$

Answer 1

We need to find value of ${lim}_{x\to 1}(\frac{1}{x^2+x-2}-\frac{x}{x^3-1})$

$={lim}_{x\to 1}(\frac{1}{(x-1)(x+2)}-\frac{x}{{(x-1)(x}^2+1+x)})={lim}_{x\to 1}\left\{\frac{1}{(x-1)}(\frac{1}{x+2}-\frac{x}{x^2+x+1})\right\}={lim}_{x\to 1}\left\{\frac{1}{(x-1)}\left(\frac{x^2+x+1-x^2-2x}{(x+2)(x^2+x+1)}\right)\right\}={lim}_{x\to 1}\left\{\frac{1}{(x-1)}\left(\frac{1-x}{(x+2)(x^2+x+1)}\right)\right\}={lim}_{x\to 1}\frac{-1}{(x+2)(x^2+x+1)}=\frac{-1}{(1+2)(1+1+1)}=-\frac{1}{9}$