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September 9th, 2017, 12:58 PM   #1
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Limit question

Hi guys,
I have the answers of the question but I didn't understand how? especially numerator part? I know the denominator is factored out and became (x+2)(x-1)

please look at the question and try to explain me.

I appreciate.
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September 9th, 2017, 12:59 PM   #2
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also can you try the same limit with (3x^2-ax+a+6)/ (x^2+x-2)
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September 9th, 2017, 01:19 PM   #3
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As the denominator is zero at the limiting value (-2) we need 2x$^2$ + ax + a + 6 = 0. We
then have the indeterminate form 0/0. So, substitute x = -2 into 2x$^2$ + ax + a + 6 = 0
and solve for a.

Apply the same method for your other problem.
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September 9th, 2017, 02:00 PM   #4
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This is super helpful. thanks Greg.
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