Question

What is `lim _(xrarr 0) (8x^2)/(cos(x)-1)` and how do you get the answer?

Answer 1

It is `-16`

Explanation:

Hence the limit `x->0` gives an indeterminate form of `0/0`
we can apply L’Hospital’s Rule hence

#lim_(x->0) (8x^2)/(cosx-1)=lim_(x->0) (d(8x^2))/dx/[((d(cosx-1))/dx)]= lim_(x->0) (16x)/(-sinx)=lim_(x->0) -16/(sinx/x)=-16/[lim_(x->0)sinx/x]= -16/1=-16#

Remember that `lim_(x->0)sinx/x=1`