Post #26

This week in calculus, we reviewed topics we should know how to do because they show up on every AP.

Limits:
LIm as x-> c The first step in finding a limit, is to plug in c. If that gives you zero at the bottom, it does not mean the limit doesn't exist. Next thing to try is to factor and cancel the top and bottom then plug in c again. If that still gives you zero, you have to use L'Hopital's rule. L'hopital's rule states if lim as x->c = 0/0, take the derivative of the top and the derivative of the bottom and plug in c. If you still get undefined, take the derivative of each again until you don't.

Example:
lim as x-> 4 3(x-4) / x^2 - 16
First step is plug in 4: 3(0) / 0 = 0/0
Factor and cancel: 3(x-4) / (x+4) (x-4) = 3/ (x+4)
Now plug in: 3/ (4+4) = 3/8, the limit is 3/8
If that would not have worked, you could have used L'hopital's rule
Take derivatve: 3/ 2x and plug in: 3/(4)(2) = 3/8

Example 2:
lim as x-> 0 4 sinx cosx - sinx / x^2
Plugging in gives you 0/0.
Factoring leaves you with 0/0.
So the only thing you can do is L'hopital's rule
Top factored: sinx (-sinx) + cosx (cosx) - cosx
Bottom factored: 2x
Still get 0 at the bottom so take derivative again
(-sin x)^2 - cos x
2 (-sin x) (cos x) + sin x
-2 sin x cos x + sin x / 2
Now plug in 0
-2 (0) (1) + (0) / 2 = 0
The limit as x-> 0 is 0.

Derivatives of integrals
To find the derivative of an integral, just plug in b the multiply that by the derivative of b.

Example:
f(x) = the integral of sin (t) on [0,x^2] Find f'(x)
sin (x^2) (2x)
f'(x) = 2x sin (x^2)

Definition of a derivatve
To recognize the definition of a derivative, the problem will look like
lim as a variable -> 0 ( + variable) - something / variable
To solve this, just take the derivative and plug in the number used.

Example: lim g -> 0 sin(pi +g) - sin (pi) / g
Take the derivative of sin (x) so cos (x)
now plug in pi - cos (pi) = -1

Example 2: lim as h -> 0 cos ( pi/4 + h) - cos (pi/4) / h
Derivative of cos (x) = -sin (x)
plug in pi / 4 = - sin (pi/4) = - square root of 2 / 2