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November 10th, 2016, 04:02 AM   #1
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Question Function Limits

I just want to ask whether I'm doing it right? If there's some part of it that I'm doing wrong, please correct me. Thanks! ^^

2.png


After that, I end up with the answer 0/-1

Is my answer acceptable or did I do something wrong?

Last edited by skipjack; November 10th, 2016 at 04:50 AM.
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November 10th, 2016, 04:14 AM   #2
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If you are asked for the limit as $y\to 2$, your method is fine, but we normally write $\frac{0}{-1}$ as $0$.
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November 10th, 2016, 04:19 AM   #3
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Aah yes~~ Thank you so much!!
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November 10th, 2016, 04:29 AM   #4
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No, that is not correct, for a couple of reasons. There is a typo on the third step- it should be "8- 4- 2- 2", not "16- 4- 2- 2". It was clearly a typo because you 8- 4- 2- 2= 0 while 16- 4- 2- 2= 8.

But your real problem is in the denominator: $\displaystyle 2(2^3)- 5(2^2)+ 5(2)- 6= 2(8 )- 5(4)+ 5(2)- 6= 16- 20+ 10- 6= 26- 26= 0$.
(You have $\displaystyle 2(2^3)= 32$ which is incorrect.)

Simply setting x= 2 gives the "indeterminate" 0/0. You will need to use another method to determine the limit. Fortunately, since both numerator and denominator are polynomials, the fact that each has value 0 when x= 2 tells you that x- 2 is a factor of each. Factor each and cancel the "x- 2"s.
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Last edited by Country Boy; November 10th, 2016 at 04:34 AM.
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November 10th, 2016, 04:42 AM   #5
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There are two errors in SlayedByMath's arithmetic. The 16 and 32 should be 8 and 16 respectively.

With those corrected, one ends up with 0/(8 - 10 + 5 - 3), i.e. 0/0. One can cancel a factor of (y-2) from the numerator and denominator to overcome that (without use of calculus) or use L'Hôpital's rule.
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