The website I am working through is: http://www.math.ucdavis.edu/~kouba/CalcO…
Not really a problem so far except for when I simplify and get 0/0, I dont understand how I continue? I have read the solution through but it is the single process of factoring out numerator and denominator I dont get.
Eg: lim x^4 - 81 / 2x^2 - 5x - 3
x ->3
I understand that it simplifies as zero no problem. Its this bit I don't get, and hopefully someone will explain how it is done:
lim (x^2 - 9) (x^2 + 9) / (x - 3) (2x + 1)
x ->3
From this point I can get the right answer fine, it is just the cross-over between these 2 points I need to understand.
Not really a problem so far except for when I simplify and get 0/0, I dont understand how I continue? I have read the solution through but it is the single process of factoring out numerator and denominator I dont get.
Eg: lim x^4 - 81 / 2x^2 - 5x - 3
x ->3
I understand that it simplifies as zero no problem. Its this bit I don't get, and hopefully someone will explain how it is done:
lim (x^2 - 9) (x^2 + 9) / (x - 3) (2x + 1)
x ->3
From this point I can get the right answer fine, it is just the cross-over between these 2 points I need to understand.
-
When you evaluate a limit and get 0/0 that means that there is a hole at the x value (in this case when x=3)
To evaluate the limit further you can cancel something out. So factor everything and you should be able to cancel everything.
x^4-81=(x^2 - 9) (x^2 + 9) and...
2x^2 - 5x - 3=(x - 3) (2x + 1)
Therefore:
lim (x^2 - 9) (x^2 + 9) / (x - 3) (2x + 1)
x ->3
lim (x-3)(x+3) (x^2 + 9) / (x - 3) (2x + 1)
x ->3
Notice that the (x-3)'s can cancel out. Therefore you have:
lim (x+3) (x^2 + 9) / (2x + 1)
x ->3
Evaluate this and you should have your answer.
To evaluate the limit further you can cancel something out. So factor everything and you should be able to cancel everything.
x^4-81=(x^2 - 9) (x^2 + 9) and...
2x^2 - 5x - 3=(x - 3) (2x + 1)
Therefore:
lim (x^2 - 9) (x^2 + 9) / (x - 3) (2x + 1)
x ->3
lim (x-3)(x+3) (x^2 + 9) / (x - 3) (2x + 1)
x ->3
Notice that the (x-3)'s can cancel out. Therefore you have:
lim (x+3) (x^2 + 9) / (2x + 1)
x ->3
Evaluate this and you should have your answer.
-
lim (x^2 - 9) (x^2 + 9) / (x - 3) (2x + 1)
x ->3
= lim [(x+3)(x-3)] * (x^2 + 9) / (x - 3) (2x + 1)
x ->3
= lim (x+3)(x^2 + 9) / [2x + 1)
x -> 3
= (3+3)*([3]^2 + 9) / (2(3) + 1)
= 108 / 7
x ->3
= lim [(x+3)(x-3)] * (x^2 + 9) / (x - 3) (2x + 1)
x ->3
= lim (x+3)(x^2 + 9) / [2x + 1)
x -> 3
= (3+3)*([3]^2 + 9) / (2(3) + 1)
= 108 / 7
-
lim [(x^2 - 9)(x^2 + 9)] / (x - 3)(2x + 1) =
x -> 3
lim [(x + 3)(x - 3)(x^2 + 9)] / (x - 3)(2x + 1) =
x -> 3
lim [(x + 3)(x^2 + 9)] / (2x + 1) = 108 / 7
x -> 3
id est
x -> 3
lim [(x + 3)(x - 3)(x^2 + 9)] / (x - 3)(2x + 1) =
x -> 3
lim [(x + 3)(x^2 + 9)] / (2x + 1) = 108 / 7
x -> 3
id est