# How to evaluate the limit of (cos x - cos 3x) / x*sin x if x-->0 ?

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### 2 Answers

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**

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**