a)
wordy solution:
by looking at that equation, you see that the largest power in the numerator is 1 (x1) and the largest power in the denominator is 2 (x2), since the denominator has the larger power, it will grow faster than the numerator as x gets bigger, so the limit as x approaches infinity of g(x) is 0.
not as wordy solution:
\(lim_{x\rightarrow\infty}\frac{x+4}{x^2+x-12}\\ =lim_{x\rightarrow\infty}\frac{x^2(\frac{1}{x}+\frac{4}{x^2})}{x^2(1+\frac{1}{x}-\frac{12}{x^2})}\\ =lim_{x\rightarrow\infty}\frac{\frac{1}{x}+\frac{4}{x^2}}{1+\frac{1}{x}-\frac{12}{x^2}}\\ =\frac{0+0}{1+0-0}=0\)
b)
horizontal asymptotes happen when the limit as x approaches infinity or negative infinity approaches some number
as found above, a horizontal asymptote is at y=0
you can also check the limit as x approaches negative infinity, but in this case it would also equal 0, so there's only one horizontal asymptote.
c)
\(\frac{x+4}{x^2+x-12}=\frac{x+4}{(x+4)(x-3)}\)
factor the numerator and denominator, if possible
you see that x+4 can be canceled out, so x=-4 must be where the hole (removeable discontinuity) is.
after canceling out the x+4 's, you're left with:
\(\frac{1}{x-3}\)
to find the coordinate of the hole, plug in x=-4 into the equation and you get y=-1/7, so, the hole is at (-4,-1/7).
to find the vertical asymptote, use the equation that's left over after the canceling and set the denominator equal to 0
x-3 = 0
x = 3
There is a vertical asymptote at x=3.
