Help with a calculus problem?

Question

What are the equations of all asymptotes of

y=(x^2-3x-the square root of x)/(x-1)

Answer 1

y = [x^2 - 3x - sqrt(x)] / (x - 1)

To solve for the vertical asymptotes; it's easy; whatever makes the denominator undefined will find it.

x - 1 = 0
x = 1; therefore, x = 1 is a vertical asymptote.

To find the horizontal asymptotes, we need to take two limits (which may or may not be the same)
the limit as the function approaches infinity
the limit as the function approaches negative infinity.

lim [x^2 - 3x - sqrt(x)] / (x - 1)
x -> infinity

Since we have an indeterminate form of the form [infinity/infinity], we apply L'Hospital's rule.

lim (2x - 3 - (1/[2sqrt(x)]))
x -> infinity

This is of the form [infinity - 3 - 0], which is infinity.
Therefore, there aren't any horizontal asymptotes.

Similarly,

lim (2x - 3 - (1/[2sqrt(x)]))
x -> -infinity

is of the form [-infinity - 3 - 0] = -infinity.
So there still aren't any horizontal asymptotes.

Answer 2

y=(x^2-3x-x^.5)/(x-1)
The curve is asymptotic only to the line x = 1.
The curve would be asymptotic to the y-axis except it becomes imaginary when x < 0

Answer 3

x=1 since curve approaches infinty at x=1

Answer 4

y=(x²-3x-the square root of x)/(x-1)
D(f)= |R - {1 and 4} or {x elements of R | x different of 4 and diferent of 1}