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 November 10th, 2016, 04:02 AM #1 Member   Joined: Aug 2016 From: South Korea Posts: 55 Thanks: 0 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.
 November 10th, 2016, 04:14 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2661 Math Focus: Mainly analysis and algebra If you are asked for the limit as $y\to 2$, your method is fine, but we normally write $\frac{0}{-1}$ as $0$. Thanks from SlayedByMath
 November 10th, 2016, 04:19 AM #3 Member   Joined: Aug 2016 From: South Korea Posts: 55 Thanks: 0 Aah yes~~ Thank you so much!!
 November 10th, 2016, 04:29 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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. Thanks from v8archie and SlayedByMath Last edited by Country Boy; November 10th, 2016 at 04:34 AM.
 November 10th, 2016, 04:42 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 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'HÃ´pital's rule.

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