DIFFERENTIAL CALCULUS OF LIMIT OF FUNCTION

What is the limit of 
(1-cosx)/x^3 as x approaches 0?
 

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  We will  differentiate the numerator f(x)=(1-cos(x)),

 then the denominator g(x)=x^3. 

That is

f'(x)=0-(-sin(x)) =sin(x) 

g'(x)= 3x^2 

Now we look at 

lim x->0 sin(x) /(3x^2). 

Since we would get 0/0, we can use L'hopital's rule again 
and consider lim x->0 f''(x) /g''(x). 

Taking the derivatives again, we get f''(x)=cos(x) and g''(x)=6x.

Now see if we can find the limit : 

lim x->0 cos(x) /6x ->1/0 ->infinity 

So,lim x->0 (1-cos(x))/x^3= infinity .

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