# calculus prob. help???

#### ladyazpy

##### New member
In, calculus, the ratio (sin x)/x comes up naturally, and the problem is to determine what the ratio approaches when x approches 0. can you guess?
(Note that the ratio is not defined when x equals 0)

A. Referring to the figure, show the following:

A sub 1 = area of triangle QAP = (sin x) / (2)

A sub 2 = area of sector OAP = (x) / (2)

A sub 3 = area of triangle OAB = (tan x) / (2)

B. Using the fact that A sub 1 < A sub 2 < A sub 3, which we can see clearly in the figure, show that

cos x < (sin x) / (x) < 1 x > 0

C. From the inequality in part B, explain how you can conclude that for x > 0, (sin x) / x approaches 0.

#### Gene

##### Senior Member
L'Hopital's rule says take the dervitive of the numerator and the denominator.

A. Not without the figure.

B. The equation should be
cos(x) < (sin x) / (x) < 1
Then you can sub cos(0) for cos(x)

c. you can't prove it -> 0 'cause it doesn't.

#### ladyazpy

##### New member
cos x < (sin x / x) < 1 x>0

this is the right equation im sorry........

#### Gene

##### Senior Member
I'm sorry too, but
cos(0)=1
L'Hopital says
lim<sub>x->0</sub>sin(x)/x =
lim<sub>x->0</sub>d(sin(x))/d(x) =
lim<sub>x->0</sub>cos(x)/1 =
1/1 = 1
1<1<1 just doesn't work.
The sandwich theorum suggested in one of your many posts relies on < also.
I don't know what to say except your source has to be as wrong with this as it is with the limit it asks you to prove.