# What is the limit of the function y given by y=(cos x-cos7x)/x, if x approaches to 0 value?

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We have to find lim x-->0 [(cos x - cos 7x)/x]

If we substitute x with 0, we get the form 0/0 which is indeterminate, this allows us to use l'Hopital's rule and substitute the numerator and the denominator with their derivatives.

We get lim x-->0 [(-sin x + 7*sin 7x)/1]

Now substitute x = 0

We get 0

**The required value of lim x-->0 [(cos x - cos 7x)/x] = 0**

We'll substitute x by the value of accumulation point, in the given expresison of function:

y = (cos 0 - cos 7*0)/0 = (1 - 1)/0 = 0/0

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

lim (cos x - cos 7x)/x = lim (cos x - cos 7x)'/(x)'

lim (cos x - cos 7x)'/(x)' = lim (-sin x + 7 sin 7x)/1

We'll substitute x by accumulation point:

lim (-sin x + 7 sin 7x) = -sin 0 + 7 sin 7*0

lim (-sin x + 7 sin 7x) = 0 - 7*0

lim (-sin x + 7 sin 7x) = 0** **

**Therefore, for x->0, the limit of the function is: lim (cos x - cos 7x)/x = 0.**