What is limit (sin x - sin 2x )/x ?x->0

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You need to replace 0 for x in limit, such that:

`lim_(x->0) (sin x - sin 2x)/x = (sin 0 - sin2*0)/0 = (0-0)/0 = 0/0`

The indetermination 0/0 requests for you to use l'Hospital's theorem, such that:

`lim_(x->0) (sin x - sin 2x)/x = lim_(x->0) ((sin x - sin 2x)')/(x') `

`lim_(x->0) ((sin x - sin 2x)')/(x') = lim_(x->0) (cos x - 2cos 2x)/1`

Replacing 0 for x yields:

`lim_(x->0) (cos x - 2cos 2x)/1 = cos 0 - 2cos 0`

Since `cos 0 = 1` yields:

`lim_(x->0) (cos x - 2cos 2x) = 1 - 2 = -1`

**Hence, evaluating the given limit, using l'Hospital's theorem, yields `lim_(x->0) (sin x - sin 2x)/x = -1` .**

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