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...
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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