How to calculate this limit? f:[0,1]->RR,f(x)=xsqrt(1-x^2) Calculate lim_(x->0)1/x^2int_0^xf(t)dt

1 Answer
mason m
Mar 25, 2017

lim_(xrarr0)1/x^2int_0^xf(t)dt=1/2

Explanation:

lim_(xrarr0)1/x^2int_0^xtsqrt(1-t^2)dt

As x approaches 0, note that there are two things happening: 1/x^2 becomes 1/0 and the integral approaches int_0^0f(t)dt, which is also 0.

Therefore the limit is in the form 0/0, which means that l'Hopital's rule applies.

We can then treat it as:

lim_(xrarr0)(int_0^xtsqrt(1-t^2)dt)/x^2

And then using l'Hopital's rule:

=lim_(xrarr0)(d/dxint_0^xtsqrt(1-t^2)dt)/(d/dxx^2)

The derivative of the numerator can be found through the Second Fundamental Theorem of Calculus. The derivative of the denominator is 2x:

=lim_(xrarr0)(xsqrt(1-x^2))/(2x)=lim_(xrarr0)sqrt(1-x^2)/2=1/2

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