# limitx->0 sin3x-5sin2x/sin6x-3sinx

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You need to evaluate limit of function `(sin 3x-5sin 2x)/(sin 6x-3sin x)` when `x-gt0` , hence you need to substitute 0 for x in equation of function such that:

`lim_(x-gt0) (sin 3x-5sin 2x)/(sin 6x-3sin x) = (sin0-5sin0)/(sin0-3sin0)`

Using `sin 0 = 0 =gt lim_(x-gt0) (sin 3x-5sin 2x)/(sin 6x-3sin x) = 0/0`

The case `0/0` is indeterminate, hence you may use l'Hospital's theorem to solve the limit:

`lim_(x-gt0) ((sin 3x-5sin 2x)')/((sin 6x-3sin x)') =lim_(x-gt0)(3cos 3x- 10 cos 2x)/(6 cos 6x - 3cos x) `

Using cos 0 = 1 yields

`lim_(x-gt0)(3cos 3x- 10 cos 2x)/(6 cos 6x - 3cos x) =(3*1 - 10*1)/(6*1 - 3*1)`

`lim_(x-gt0)(3cos 3x- 10 cos 2x)/(6 cos 6x - 3cos x) = -7/3`

**Hence, evaluating the limit of function yields `lim_(x-gt0) (sin 3x-5sin 2x)/(sin 6x-3sin x) = -7/3` .**