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#.
