Determine the limit of the function (sin5x-sin3x)/x, x-->0
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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
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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.
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