Find the following limit as x->0

[thenewbosco]

find the following limit as x-->0

Hello I am trying to find the following limit as x-->0. I have tried using l'hopital's rule but all it produces is a more complex thing i still can't take the limit of. any help?
$$\frac{sin x - x}{sin^3(x)}$$
I can see this limit exists on the graph i just don't know how to go about solving it.

 

[arildno]

Take L'hopital once.
Then multiply your expression with $$1=\frac{\cos(x)+1}{\cos(x)+1}$$

 

[thenewbosco]

Thanks this worked. One question about this is how did you know to multiply by this? How would one know to do this when faced with similar situations

 

[Hurkyl]

You could have just done L'Hôpital again.

arildno just factored the denominator (after converting sin² to cos²).

 

[arildno]

Thanks this worked. One question about this is how did you know to multiply by this? How would one know to do this when faced with similar situations

Just practice&patience.
To think of simple tricks like this comes naturally to you if you have worked assiduously with your maths earlier.

 

[arildno]

You could have just done L'Hôpital again.

arildno just factored the denominator (after converting sin² to cos²).

Factorization is more fun.