# a.p. calculus help please!

posted by
**teamcaridee**
.

My a.p. calculus teacher doesn't know how to do this problem (like many others) so he's giving us 5 points extra credit on our test if we can figure it out, but I don't know how to do it:

It's as the lim-->infinity

(x+sinx)/(x+cosx)

We know the answer is 1 because it's in the back of the book but do not know how to get that answer. Yeah he ticks me off with not knowing his own material but can anyone help me?

SORRY TO BOTHER YOU CAN YOU HELP ME WITH THIS ANSWER AND I'LL ASK AROUND FOR YOUR QUESTION TILL WE GET IT.

How do i solve inequalitites with three variables. I still can't figure it outplease help.

x+y=1

y-z=-3

2x+3y+z=1

Easy. Divide the numerator and denomiter by 1/x

(1+ sinx/x) / (1 + cosx /x)

Now take the limits as x >> inf

the sinx/x goes to zero, as does cosx/x

you are left with 1/1

You need to do a matrice I think.

x+y=1

y-z=-3

2x+3y+z=1

1 1 0 * 1 1st eq'n, 1x, 1y, 0z =1

0 1 -1 * -3 2nd eq'n, 0x, 1y,-1z=-3

2 3 1 * 1 3rd eq'n, 2x, 3y, 1z=1

you need a 1 in a diagonal and all the rest to be zeros (to the left of the *)

multiply or divide using rows (R1, R2, or R3)

to get a zero in a spot, multiply that row with a formula

1st one: -2R1+R3

multiply by the negative number of the row that has the new "1" spot then add it to the row you are changing to a zero. go all the way across.

2nd one: -R2+R1 and -R2+R3

3rd one: first, divide only by 2 to get the last "1" in the diagonal, then, -R3+R1 and R3+ R2

you get

1 0 0 * 3 x=3

0 1 0 * -2 y=-2

0 0 1 * 1 z=1

ln(x)^2/x need help with this problem

how do you solve this problem

Consider F(x)= (int[1+(cos^4(t))]dt,t,0,x)

Calculate to correctly to three decimal places :

F(pie)