Rewrite fraction to calculate limit I am practising finding limits. However, I can't seem to figure out this one. 
$$f(x) = \frac{x^3 + 4x - 5}{x^2-1}\text{ as $x$ goes to $1$}$$
I understand I have to rewrite the fraction somehow for the denominator not to equal 0, but I don't know where to start.
 A: Using the Euclidean division of $x^3+4x-5$ by $x-1$ we get
$$f(x) = \frac{x^3 + 4x - 5}{x^2-1}=\frac{(x-1)(x^2+x+5)}{(x-1)(x+1)}$$
A: One idea is to use polynomial long division.
The idea is to note that you have a cubic divided by a quadratic, so the degree of the numerator is greater by $1.$ Consequently, we can conclude that $$f(x)=ax+b+\frac{cx+d}{x^2-1}$$ for some constants $a,b,c,d,$ where the linear numerator $cx+d$ is to allow for the fact that there may be a remainder term, which is necessarily of lower degree than the denominator.
Multiplying both sides of this equation by $x^2-1$--which is non-$0$ for $x$ sufficiently close (but not equal) to $1$--we obtain $$x^3+4x-5=(ax+b)(x^2-1)+cx+d.$$ Expand the product on the right-hand side to give yourself a system of equations. Solve for $a,b,c,d.$
Once you've found these, the rest should fall right out of your usual limit manipulations.
A: $$x^3+4x-5=(x-1)(x^2+x+5)$$
$$x^2-1=(x+1)(x-1)$$

So you your fraction is :
$$\frac{(x-1)(x^2+x+5)}{(x+1)(x-1)}= \frac{(x^2+x+5)}{x+1}$$

A: You might also try L'Hôpital's rule, because in 
$$
\frac{x^3 + 4x - 5}{x^2-1}
$$
the individual limits for $x \to 1$ of nominator and denominator go to $0$ each.
The derivative  operation will reduce the denominator from quadratic to linear and you end up with an easier term:
$$
\lim_{x\to 1} \frac{x^3 + 4x - 5}{x^2-1}
= \lim_{x\to 1} \frac{3 x^2 + 4}{2 x}
= \frac{7}{2}
$$
