Vertical asymptotes exist if function has discontinuity so we have to find domain:
5-x !=0 => x !=5 => D_f=(-oo,5)uu(5,+oo)
Function has a discontinuity at x=5, vertical asymptote exists and it is: x=5.
We will examine what happens left and right of the asymptote:
lim_(x->5-epsilon) (x+5)/(x-5) = lim_(epsilon->0) (5-epsilon+5)/(5-epsilon-5) = lim_(epsilon->0) (10-epsilon)/-epsilon=
=-lim_(epsilon->0) 10/epsilon + lim_(epsilon->0) (-epsilon)/-epsilon=-oo+1=-oo
lim_(x->5+epsilon) (x+5)/(x-5) = lim_(epsilon->0) (5+epsilon+5)/(5+epsilon-5) = lim_(epsilon->0) (10+epsilon)/(+epsilon)=
=lim_(epsilon->0) 10/epsilon + lim_(epsilon->0) (epsilon)/epsilon=+oo+1=+oo
Horizontal asymptotes exist if exists limit lim_(x->oo)f(x).
We will examine two limits: when x->+oo and when x->-oo because there might exist different asymptotes in these extreme cases.
lim_(x->-oo) (x+5)/(5-x) = lim_(x->-oo) (1+5/x)/(5/x-1)=1/-1=-1
lim_(x->+oo) (x+5)/(5-x) = lim_(x->+oo) (1+5/x)/(5/x-1)=1/-1=-1
Note: 5/x->0 when x->-oo or when x->+oo.
We've found the limits (and they are the same), so R(x) have horizontal asymptote: y=-1.
If horizontal asymptote exists, oblique asymptotes don't exist and vice versa. So we won't examine oblique asymptotes.
Finally, function R(x) has vertical asymptote x=5 and horizontal asymptote y=-1.
