Question

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

Answer 1

`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`