How do you find the limit of #(x-2)/(x^2+x-6)# x as it approaches pi/2?

1 Answer
Jim H
Sep 14, 2015

Because #pi/2# does not make the denominator #0#, you can find the limit by substitution, but the arithmetic is simpler if we reduce first.

Explanation:

#(x-2)/(x^2+x-6) = (x-2)/((x+3)(x-2)) = 1/(x+3)#.

So,

#lim_(xrarrpi/2) (x-2)/(x^2+x-6) = lim_(xrarrpi/2)1/(x+3) = 1/(pi/2+3) = 2/(pi+6)#

Direct substitution will get the same answer, but making it look like this requires some algebra.

#(pi/2-2)/(pi^2/4+pi/2-6) = (2pi-8)/(pi^2+2pi-24)# #" "# (multiply by #4/4#)

# = (2(pi-4))/((pi+6)(pi-4))#

# = 2/(pi+6)#

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