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#