Let $a\in\mathbb{R}$. Define $h(x)=\frac{1}{x^2-1}-\frac{a}{x^3-1}$. Find all values of $a$ for which $\lim_{x\to 1}h(x)$ exists.
attempt
$$h(x)=\frac{1}{x^2-1}-\frac{a}{x^3-1}=\frac{1}{x-1}\Big(\frac{1}{x+1}-\frac{a}{x^2+x+1}\Big)$$Denote: $\lim_{x\to 1}h(x)=L$ $$A=\lim_{x\to 1}[(x-1)h(x)]=0\cdot L=0$$$$B=\lim_{x\to 1}\Big(\frac{1}{x+1}-\frac{a}{x^2+x+1}\Big)=\frac{1}{2}-\frac{a}{3}$$Therefore,$$A=B\iff\frac{1}{2}-\frac{a}{3}=0\iff a=\frac{3}{2}$$So it is necessary that $a=\frac{3}{2}$, though I'm struggling to show that it's sufficient.
You can show sufficiency by saying: if $a$ is not equal to $3/2$ then the $(x-1)$ term does not cancel out.