If you put `pi` instead of `x` you get indefinite expression `0/0` . So you need to use L'Hospitals rule.` `` `

`lim_(x->pi)(1-cos8x)/(1-cos2x)=lim_(x->pi)((1-cos8x)')/((1-cos2x)')=lim_(x->pi)4cdot(sin8x)/(sin2x)`

` `Again if we put `pi` instead of `x` we get `0/0` so we again apply L'Hospitals rule.

`lim_(x->pi)4cdot(sin8x)/(sin2x)=lim_(x->pi)4cdot((sin8x)')/((sin2x)')=lim_(x->pi)4cdot4cdot(cos8x)/(cos2x)=`

`16cdot (cos8pi)/(cos2pi)=16`

**Hence your limit is** `lim_(x->pi)(1-cos8x)/(1-cos2x)=16`

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