As #x# approaches #3# from the right, the numerator is a positive number and the denominator is also positive.
Numerator #rarr# a positive
Denominator #rarr# #0# through positive values,
So the ratio increases without bound.
We write #lim_(xrarr3^+) 2/(x-3) = oo#.
Although not in widespread use (I think), it may be helpful to write:
#lim_(xrarr3^+) 2/(x-3)# has form #+/0^+#, which leads to #lim_(xrarr3^+) 2/(x-3) = oo#.
Bonus
Similar reason leads to
#lim_(xrarr3^+) (-5)/(x-3)# has form #-/0^+#, which leads to #lim_(xrarr3^+) (-5)/(x-3) = -oo#
#lim_(xrarr3^-) 2/(x-3)# has form #+/0^-#, which leads to #lim_(xrarr3^-) 2/(x-3) = - oo#
#lim_(xrarr3^+) (x-1)/(x-3)# has form #+/0^+#, which leads to #lim_(xrarr3^+) (x-1)/(x-3) = oo#
#lim_(xrarr3^+) (x-7)/(x-3)# has form #-/0^+#, which leads to #lim_(xrarr3^+) (x-7)/(x-3) = -oo#
