Rewrite fraction to calculate limitI am practising finding limits. However, I can't seem to figure...

Using the Euclidean division of $x^3+4x-5$ by $x-1$ we get
$f(x)=\frac{x^3+4x-5}{x^2-1}=\frac{(x-1)(x^2+x+5)}{(x-1)(x+1)}$

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.