# 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#

We can then treat it as:

And then using l'Hopital's rule:

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To calculate the limit, first integrate the function ( f(t) ) from 0 to ( x ) and then divide the result by ( x^2 ). So, the steps are:

- Integrate ( f(t) = x\sqrt{1-x^2} ) with respect to ( t ) from 0 to ( x ).
- Divide the result by ( x^2 ).
- Finally, take the limit as ( x ) approaches 0.

Now, let's perform the calculation.

- Integrate ( f(t) = x\sqrt{1-x^2} ) with respect to ( t ) from 0 to ( x ):

[ \int_{0}^{x} f(t) dt = \int_{0}^{x} t\sqrt{1-t^2} dt ]

- Perform the integration:

[ = \frac{-1}{3}(1-t^2)^{\frac{3}{2}} \Bigg|_{0}^{x} ]

[ = \frac{-1}{3}(1-x^2)^{\frac{3}{2}} - \frac{-1}{3}(1-0^2)^{\frac{3}{2}} ]

[ = \frac{-1}{3}(1-x^2)^{\frac{3}{2}} - \frac{-1}{3} ]

[ = \frac{-1}{3}(1-x^2)^{\frac{3}{2}} + \frac{1}{3} ]

- Divide the result by ( x^2 ):

[ \frac{1}{x^2} \int_{0}^{x} f(t) dt = \frac{1}{x^2} \left( \frac{-1}{3}(1-x^2)^{\frac{3}{2}} + \frac{1}{3} \right) ]

[ = \frac{-1}{3x^2}(1-x^2)^{\frac{3}{2}} + \frac{1}{3x^2} ]

- Finally, take the limit as ( x ) approaches 0:

[ \lim_{x \to 0} \frac{1}{x^2} \int_{0}^{x} f(t) dt = \lim_{x \to 0} \left( \frac{-1}{3x^2}(1-x^2)^{\frac{3}{2}} + \frac{1}{3x^2} \right) ]

[ = \frac{-1}{3}(1-0)^{\frac{3}{2}} + \frac{1}{3} ]

[ = -\frac{1}{3} + \frac{1}{3} ]

[ = 0 ]

So, the limit as ( x ) approaches 0 of ( \frac{1}{x^2} \int_{0}^{x} f(t) dt ) is 0.

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