Question

compute using algebraic approach
lim as x---> positive infinity of 4-x/ x^2+5x-14 ????

Answer 1

pull out x as common factor from numerator and x^2 from denominator, your answer should be zero

Answer 2

do you mean factor the numerator and denominator? top would be (x+2)(x-2) and bottom would be (x-2)(x+7) ??

Answer 3

well you have the numerator as x-4 or is it x^2-4. Please type the question correctly

Answer 4

oh sorry the top is 4-x^2

Answer 5

ok, then we have a different answer. so yes you will factor that out as x+2 * x-2 and what you did with denominator was good x+7 * x-2

Answer 6

somethings will cancel out

Answer 7

yes and then we cancel out the x-2

Answer 8

and you should be on your way to the answer, you can also pull out X as common factor after cancelling terms and then your answer should be 1

Answer 9

what do i cancel with x+2/ x+7.. i'll be left out with 2/7 ?

Answer 10

what's the infinity's function here??

Answer 11

since we cant plug for x ?

Answer 12

when you are left with x+2 / x+7 , if you put in limits it will be inf / inf , so you can evaluate limits but we can do this

Answer 13

makes sense ? i have shown a short cut, alternatively since x+2 / x+7 is inf / inf form you use L HOSPITAL rule, but i dont see the need here

Answer 14

i understand you factored x out, but why you divide by x ?

Answer 15

ok does 7 or 2 have an x term with them ? the ansWER IS NO. so if you want a common factor from say x+2 , i.e. pull out x as common factor from X +2 you gotta divide 2 by x , because 2 DOES NOT have x term. same deal with x+7

Answer 16

the end didn't make sense for me. but thanks alot