1 answer

How do you find the limit of #(x+3)/(x-3)# as x approaches 3+?

Question:

How do you find the limit of #(x+3)/(x-3)# as x approaches 3+?

Answers

As #xrarr3^+#, the numerator #x+3# #rarr 6#.

The denominator #x-3# #rarr 0#.

So the fraction is either increasing without bound (#rarroo#) or decreasing without bound (#rarr -oo#)

As #xrarr3^+#, we know that #x > 3# so #x-3 > 3-3 =0#. That is

As #xrarr3^+#, #x-3# #rarr 0# through positive values.

(As #xrarr3^+#, #x-3# #rarr 0^+# or #lim_(xrarr3^+)(x-3) = 0^+#)

The numerator is going to a positive and the denominator is going to 0, through positives, so the fraction is increasing without bound.

#lim_(xrarr3^+)(x+3)/(x-3) = oo#

.

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