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

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Answer 1 · 2015-09-14T20:48:21

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.

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

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 $4/4$ )

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

$= 2/(pi+6)$

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