We have to find the value of lim x--> 0[ (sin 5x - sin 3x)/x]

if we substitute x = 0, we get the form 0/0, which allows us to use the l'Hopital's rule and substitute the numerator and the denominator by their derivatives.

=> lim x--> 0[ 5*cos 5x - 3*cos 3x]

substitute x = 0

=> 5*1 - 3*1

=> 2

**The required value of lim x-> 0[ (sin 5x - sin 3x)/x] = 2**

First, we'll verify if we'll get an indetermination by substituting x by the value of the accumulation point.

lim (sin5x-sin3x)/x = (0 - 0)/0 = 0/0

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

lim (sin5x-sin3x)/x = lim (sin5x-sin3x)'/(x)'

lim (sin5x-sin3x)'/(x)' = lim (5cos 5x- 3cos 3x)/1

We'll substitute x by accumulation point:

lim (5cos 5x- 3cos 3x) = 5cos5*0 - 3cos 3*0

lim (5cos 5x- 3cos 3x) = 5*1 - 3*1

lim (5cos 5x- 3cos 3x) = 2

**For x->0, lim (sin5x-sin3x)/x = 2.**