What is the slant asymptote of the function y=x^2/(x+1)?

We'll recall the equation of the slant
asymptote:

y = mx + n

We must
identify the coefficients m and n to determine the equation of the slant
asymptote.

m = lim f(x)/x, if x approaches to
+infinite

Let f(x)=y

m = lim
x^2/x*(x+1)

We'll remove the brackets from
denominator:

lim x^2/x*(x+1) = lim x^2/(x^2 +
x)

We'll force the factor x^2 at
denominator:

lim x^2/x^2*(1 + 1/x) = lim 1/(1 +
1/x)

lim 1/(1 + 1/x) = lim 1/(1 + lim
1/x)

lim 1/(1 + lim 1/x) = 1/(1+0) =
1

Since m = 1, we may calculate
n:

n = lim [f(x) - mx] = lim [x^2/(x+1) -
x]

lim [x^2/(x+1) - x] = lim (x^2 - x^2 -
x)/(x+1)

lim (- x)/(x+1) = -1/1 =
-1

The equation of the slant asymptote, if x
approaches to + infinite and - infinite, is y = x -
1.