# Finding the limit of trignometric functions

1. Sep 29, 2010

1. The problem statement, all variables and given/known data

Find the Limit: lim h->0 of ( h sin h)/ ( 1-cos h)

2. Relevant equations

lim h->0 of (sin theta)/ theta = 1
lim h->0 of (cos(h)-1)/h=0

3. The attempt at a solution
lim h->0 (h/1-cos h)*(sin h) * (1+cos h/ 1 +cos h)

= lim h->0 (h(1+cos h) * sin h)/(1-cos^2 h)

= lim h->0 (h(1+cos h))/sin h

That's how far I got, I have no idea how to continue.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 29, 2010

### Staff: Mentor

You're almost there!
$$=\lim_{h \to 0} \frac{h}{sin(h)}\cdot (1 + cos(h))$$

You can break up the limit of a product to the product of limits, provided that both limits exist. Hint: look at your relevant equations.

3. Sep 29, 2010

Okay, so I split them up, but would the equation: lim h->0 of sin h/ h = 1 be the same as the lim h->0 of h/sin h?? how does that work?

and therefore be: 1 * lim h->0 (1+ cos h)
= 1 * 2 = 2?

4. Sep 29, 2010

### JonF

Yup, I get 2 also.

If you know limh→0 h/sinh = 1 then you know that
limh→0 (sinh/h)-1 = 1.

But (sinh/h)-1 = 1 ⇒ (sinh/h)= 1

Edit: You don’t know l’hopital’s rule right? If you do this is a rather round-about way to do this problem.

Last edited: Sep 29, 2010
5. Sep 29, 2010

### Staff: Mentor

Agreed.
Not sure what the missing symbol is - implies?
But sinh/h $\neq$ 1. The limit as h -> 0 is 1, though.
But if the OP doesn't know L'Hopital's Rule, this is the way to go.

6. Sep 29, 2010