# 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.

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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

Staff Emeritus
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You could have just done L'Hôpital again.

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

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thenewbosco said:
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.