On
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.
On
$$x^3+4x-5=(x-1)(x^2+x+5)$$
$$x^2-1=(x+1)(x-1)$$
So you your fraction is :
$$\frac{(x-1)(x^2+x+5)}{(x+1)(x-1)}= \frac{(x^2+x+5)}{x+1}$$
On
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: $$ \lim_{x\to 1} \frac{x^3 + 4x - 5}{x^2-1} = \lim_{x\to 1} \frac{3 x^2 + 4}{2 x} = \frac{7}{2} $$
Using the Euclidean division of $x^3+4x-5$ by $x-1$ we get
$$f(x) = \frac{x^3 + 4x - 5}{x^2-1}=\frac{(x-1)(x^2+x+5)}{(x-1)(x+1)}$$