# indeterminationif evaluate (sin2x-sin6x)/x, for x-->0, the result is 0/0. what is to do further? is possible to solve the limit?

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### 2 Answers

We need the value of lim x--> 0[ (sin2x-sin6x)/x]

substituting x=0 gives the form 0/0, which is indeterminate. So use the l'Hopital's rule and substitute the numerator and the denominator by their derivatives.

=> lim x--> 0[ (2*cos 2x - 6*cos 6x)/1]

substitute x = 0

=> 2*1 - 6*1 = 2 - 6 = -4

**The required value of the limit is -4**

### Since we've get an indetermination, we'll apply l'Hospital rule:

lim (sin2x-sin6x)/x = lim (sin2x-sin6x)'/(x)'

lim (sin2x-sin6x)'/(x)' = lim (2cos 2x- 6cos 6x)/1

We'll substitute x by accumulation point:

lim (2cos 2x- 6cos 6x) = 2cos 2*0- 6cos 6*0

lim (2cos 2x- 6cos 6x) = 2*1 - 6*1

lim (2cos 2x- 6cos 6x) = -4** **

**Therefore, for x->0, the limit of the function exists and it is finite, it's value being: lim (sin2x-sin6x)/x = -4.**