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?

Since the numerator and denominator is zero at $1$, let's factor out $(x-1)$ from both of them to get an idea how the function behaves around $1$.
The fraction equals $\dfrac{(3x^3-5x^2-5x-5)(x-1)}{(x^2-1)(x-1)}=\dfrac{3x^3-5x^2-5x-5}{x^2-1}$.
At $x=1$, the numerator equals $-12$. So for values around and very close to $1$, the numerator stays near $-12$.
The denominator however, is negative for $x<1$ and is positive for $x>1$. Thus, as $x$ approaches $1$ from the left, $x^2-1$ takes on values like $-0.1,-0.01,-0.001,\ldots$ while the numerator remains close to $-12$. Hence, the fraction is positive and becomes arbitrarily large as $x\to 1^{-}$. 
Similarly, as $x\to 1^{+}$, the denominator is positive and becomes small while the numerator remains near $-12$ so that your expression here approaches $-\infty$.
