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

Answer 1

It is #-16#

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#

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Answer 2

Charles Arocha

The limit of (8x^2)/(cos(x)-1) as x approaches 0 can be found using L'Hôpital's rule. By differentiating the numerator and denominator separately, we get (16x)/(sin(x)). Evaluating this expression at x=0, we find that the limit is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.