difficult calculus problem!!?

Question

The following informations is given about a function f(x) which is defined and has first and second derivatives at all real numbers except x= 3.

lim f(x) = −1;
x→ −∞

lim f(x) = ∞; f′(−0.5)=0, f"(−0.5)=1.
x→ 3+

We conclude that f(x) has
a horizontal asymptote (to the left) y=?
a vertical asymptote x=?
a local extremum at x=?
Use the second derivative test to describe the extremum. Write LMAX for relative maximum, LMIN for relative minimum?

Answer 1

Since lim f(x) = −1;
x→ −∞

it follows f MAY have a horizontal asymptote to the left. The existence of the limit doesn't necessarily imply the existence of this asymptote, because the function may oscillate around -1. For example, lim x -> -oo sin(x)/x - 1 = -1, but this function doesn't have an asymptote, because it's graph bounces up and down around -1 as x goes to - oo.

Since
lim f(x) = ∞;
x→ 3+

it does follow f has a vertical asymptote to the right of the axis x = 3. Because to each x there corresponds only one f(x).

Since f′(−0.5)=0 and f"(−0.5)=1, that is, first derivative vanishes at -0.5 and second derivative is positive at -0.5, it follows f has a LMIN at x = -0.5.