How do you find the limit of #(x^2-9)/(x^2+2x-3)# as x approaches #1^+#?

1 Answer
Jim H
Nov 8, 2016

#lim_(xrarr1^-)(x^2-9)/(x^2+2x-3) = -oo#

Explanation:

When we try to evaluate #lim_(xrarr1^-)(x^2-9)/(x^2+2x-3) # by substitution we get a non-#0# number over #0#. This tells us that we have some kind of infinite limits at #1#.

To determine whether the ratio is increasing or decreasing without bound (going to #oo# or #-oo#) we need to investigate the signs of the factors.

#lim_(xrarr1^-)(x^2-9)/(x^2+2x-3) = lim_(xrarr1^-)(x^2-9)/((x+3)(x-1))#

As #xrarr1^+#,

#x^2-9 rarr -8#, #" "# (The numerator is negative.)
#x+3rarr4#, #" "# (This factor is positive.)
#x-1rarr0^+#, #" "# (This factor is positive.)

The ratio is decreasing without bound.

We write:

#lim_(xrarr1^-)(x^2-9)/(x^2+2x-3) = -oo#

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