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?
 A: 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 $-\infty$.
A: You've learned how to use features of polynomials and rational functions -- e.g. things like finding the zeroes and stuff -- to help you graph them, right?
You can use the same idea in reverse: what feature of a rational function corresponds to a pole? (i.e. to the graphical feature you've observed)
A: Hints:
$$3x^4-8x^3+5=(x-1)(3x^3-5x^2-5x-5)$$
$$x^3-x^2-x+1=(x-1)(x^2-1)=\ldots$$
A: We have
$$
\frac{3x^4-8x^3+5}{x^3-x^2-x+1} = 3x - 5 + \frac{4}{x+1} - \frac{6}{x-1},
$$
which can be found by using polynomial long division and partial fraction decomposition. From this form the limits $x\to 1^+$ and $x\to 1^-$ are easy to compute without looking at any plots.
