Finding one sided limits algebraically I was wondering what the best method was for proving this limit algebraically:
$$\lim_{x \to 1}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}$$
I know the answer to this question is ;
$$\lim_{x \to 1^+}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={-\infty}$$ 
$$\lim_{x \to 1^-}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={+\infty}$$
The only way I can solve this is using the graph of the function as the limit becomes very apparent. What is the best way to answer this algebraically?

Hints:
$$3x^4-8x^3+5=(x-1)(3x^3-5x^2-5x-5)$$
$$x^3-x^2-x+1=(x-1)(x^2-1)=\ldots$$
