Calculus help????????

f(x)= (2x − 2) / (x² + x − 2)
I have vertical asymptote @ x=-2
horizontal asymptote @ y=0

how can i justify my answer using limits?

f(x)=(2x-2)/(x²+x-2)

the vertical asymptotes occur when
the denominator is 0:

x²+x-2=0
(x+2)(x-1)=0
x=-2 and x=1

But notice that x=1 is a HOLE!
by plugging x=1 you'll get a 0/0 form.
so the only vertical asymptote is x=-2
--------------------
for the horizontal asymptotes,you need
to find the limit as x approaches infinity and minus infinity.

lim (2x-2)/(x²+x-2)
x->∞

lim 2(x-1)/(x-1)(x+2)
x->∞

lim 2/(x+2)
x->∞

as lim x->∞ then 2/(∞+2) =2/∞ = 0

for x->-∞ the limit is also 0

so the horizontal asymptote is y=0

You dont really use limits to find a vertical asymptote although you could take the limit as x-->-2+ and -2-

to figure out whether itll be tending towards negative or positive infinity.

for horizontal asymptote you take the limit as x-->∞ and -∞

but in this case since its a polynomial, we can only go with 1 as it will be the same on the other side.

lim x-->∞ (2x/x^2)
lim x-->∞ (2/x) = 0

f(x) = 2(x - 1)/(x + 2)(x - 1) = 2/(x + 2)

Hence, vertical asymptote at x = -2 and horizontal asymptote at y = 0

as x ---> -2 from the negative side, x + 2 ---> -0.00000000.....2/(x + 2) ---> - ∞

as x ---> -2 from the positive side, x + 2 ---> 0.000000000....2/(x + 2) ---> + ∞

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