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Saturday, January 26, 2008
Calculus One
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5 comments:
Find the number a such that the limit exist?
lim (5x^2+ax+a+5)/(x^2-x-6)
x->-2
The denominator can be factor as (x-3)(x+2).
This means that the numerator should preferably contain a factor of (x+2) so we can cancel (x+2) and limit will exist when x approaches to -2.
So, we have
(Eq#1):
5x^2 + ax + (a+5) = 0
Substitute x = -2 in (Eq#1).
20 - 2a + a + 5 = 0
a = 25
Which gives us
5x^2+25x+30= (5x+15)(x+2)
So the new function
lim (5x^2+ax+a+5)/(x^2-x-6)
x->-2
becomes lim (5x+15)(x+2)/(x-3)(x+2)
x>-2
= lim (5x+15)/(x-3) = -1
x>-2
Therefore, a = 25.
derivative of tan(t^2 - t)
derivative of tan(t^2 - t) =
(2t-1)[sec^2(t^2 - t)]
Note: You must use chain rule.
what is the domain of
1/square root of (x+6)
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