Solving Simple Trig Limit Homework

[wvcaudill2]

Homework Statement

lim _x->0 (1-cosx)/sinx

Homework Equations

(sinx)/(1+cosx) = (1-cosx)/(sinx)

sinx/x = 1

The Attempt at a Solution

Well, I know I need to change my original limit, as I cannot let the denominator be 0. I usually try to factor out things, but there is nothing to factor out here. I'm not really sure what to do.

 

[JeSuisConf]

L'Hopital's rule might work well here. Have a look at that.

 

[wvcaudill2]

L'Hopital's rule might work well here. Have a look at that.

Im afraid I don't know how to use this. I am in Calculus AB, so it is not something we learn. Any other ideas?

We usually factor, rationalize, etc. to find limits.

 

[Strants]

If

\frac{1 - cos x}{sin x} = \frac{sin x}{1 + cos x}

then, won't the limits of the two equations be equal?

 

[JeSuisConf]

Im afraid I don't know how to use this. I am in Calculus AB, so it is not something we learn. Any other ideas?

We usually factor, rationalize, etc. to find limits.

L'Hoptial's rule is extremely simple, and it makes solving the problem very obvious in this case. Anyways, with the equation you gave, the two limits will be the same as Strants said. You don't need any fancy work to evaluate that limit if you use that equation, because you can evaluate it exactly at the point x=0.

 

[wvcaudill2]

If

\frac{1 - cos x}{sin x} = \frac{sin x}{1 + cos x}

then, won't the limits of the two equations be equal?

I would think so, but when you graph this, it looks like there could be many limits.

 

[wvcaudill2]

L'Hoptial's rule is extremely simple, and it makes solving the problem very obvious in this case. Anyways, with the equation you gave, the two limits will be the same as Strants said. You don't need any fancy work to evaluate that limit if you use that equation, because you can evaluate it exactly at the point x=0.

ok, thanks!