Find the limit its calculus please help :) thanks everyone seriously thank you.
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Find the limit its calculus please help :) thanks everyone seriously thank you.

[From: ] [author: ] [Date: 11-10-11] [Hit: ]
also, IF anyone knows this... that would help too........
23. lim (x -> 1-) x^2 - 4x + 3 / x^3 - x

24. lim (x-> 4+) x^2 - 6x + 8/ x^3 - 4x

find the limit if it exists.

39. lim (x-> 7+) -7sqrt(x-7)^3 / x-7


also, IF anyone knows this... that would help too..

40. Use the Intermediate Value Theorem to prove that 6x^3 + 5x^2 + 4x + 7 = 0
has a solution between -2 and -1.


sorry if this is a lot i just have a lot of questions thank so much for everyone that has been helping me with these no words can describe how thankful i am for you guys :) so thank you once again

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23. lim (x -> 1-) x^2 - 4x + 3 / x^3 - x = lim (x -> 1-) [(x-1)*(x-3)] / [x*(x-1)*(x+1)]
simplify (x-1)'s in the numerator and denominator
= lim (x -> 1-) (x-3)/(x*(x+1))
plug 1 for x
= -2/(1*2) = -1

24. plug 4 in the function, lim=0

39. lim (x-> 7+) -7sqrt(x-7)^3 / x-7 = lim (x-> 7+) -7*(x-7)^(3/2) / x-7
simplify (x-7)^(3/2) with (x-7) in the denominator, you will have
lim (x-> 7+) -7sqrt(x-7)
now plug x=7, lim = 0

40. It's a polynomial function and polynomial functions are continuous everywhere.
f(-2) = -48 + 20 - 8 + 7 = -29
f(-1) = -6 + 5 - 4 + 7 = 2
since the function is continuous, when going from -2 to -1 it must cut 0, so there's a solution between -2 and -1

-
Limits. Why couldn't I remember
anything about limits?
Limits. That was the week
Aaron got his hair cut.
Oh, God, he looked so cute.
OK, focus, Cady.
What was on the board
behind Aaron's head?
If the limit never approaches
anything...
The limit does not exist.
The limit does not exist!
1
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