How to evaluate the limit of (cos x - cos 3x) / x*sin x if x-->0 ?
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We have to determine the value of lim x--> 0[(cos x - cos 3x) / x*sin x
If we substitute x = 0, we get
(1- 1) / 0 = 0/0
As this is an indeterminate form we use l'Hopital's rule and replace the numerator and denominator by their derivatives.
=> lim x--> 0[ -sin x + 3 sin 3x] / (sin x + x*cos x)
substitute x = 0
=> (0 + 0) / (0 + 0)
Again substitute the derivatives
=> lim x--> 0[ -cos x + 9 cos 3x] / ( cos x + cos x - x*sin x)
substitute x = 0
=> (-1 + 9) / (1 + 1)
=> 8/2
=> 4
The required value of lim x--> 0[(cos x - cos 3x) / x*sin x is 4
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Since the trigonometric functions from numerator are matching, we'll transform the difference into a product:
cos x - cos 3x = 2[sin (x+3x)/2]*[sin(3x-x)/2]
cos x - cos 3x = 2 sin2x *sin x
We'll re-write the fraction:
(cos x - cos 3x) / x*sin x = 2 sin2x *sin x/x*sin x
We'll simplify and we'll get:
2 sin2x *sin x/x*sin x = 2 sin2x/x
Now, we'll evaluate the limit:
lim (cos x - cos 3x) / x*sin x = lim 2 sin2x/x
We'll create the remarcable limit:
lim 2 sin2x/x = 2 lim (sin2x/2x)*2
2 lim (sin2x/2x)*2 = 4 lim (sin2x/2x)
But lim (sin2x/2x) =1, if x -> 0
lim (cos x - cos 3x) / x*sin x = 4
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