Question

We have `f:[0,2]->RR,f(x)=sqrt(4-x^2)`.How to solve this limit?`lim_(x->0)1/x^2int_0^xtf(t)dt`

Answer 1

`1.`

Explanation:

We have, `inttf(t)dt=inttsqrt(4-t^2)dt=-1/2int(-2t)sqrt(4-t^2)dt.`

Subst. `4-t^2=u :. -2tdt=du.`

`:. int tf(t)dt=-1/2intu^(1/2)du=-1/2*u^(1/2+1)/(1/2+1)`

`=-1/2*2/3*u^(3/2)=-1/3(4-t^2)^(3/2)+c.`

`rArr int_0^x tf(t)dt=-1/3[(4-t^2)^(3/2)]_0^x`

`=-1/3(4-x^2)^(3/2)-{-1/3(4-0)^(3/2)}`

`=1/3(2^2)^(3/2)-1/3(4-x^2)^(3/2)=1/3{8-(4-x^2)^(3/2)}`

`"Therefore, the Limit="lim_(x to 0) 1/x^2int_0^x tf(t)dt`

`=lim_(x to 0) {8-(4-x^2)^(3/2)}/(3(x^2))`

Let, `4-x^2=y," so that, "x^2=4-y`.

Also, as `x to 0, y to 4.`

`:.," The Limit="lim_(yto4) (8-y^(3/2))/(3(4-y))=lim_(yto4){y^(3/2)-4^(3/2)}/{3(y-4)}`

`=1/3*3/2*4^(3/2-1)=1/2*4^(1/2)=1/2*2=1.`

The Final Limit has been evaluated using the following

Standar Limit : `lim_(x to a) (x^n-a^n)/(x-a)=n*a^(n-1); n,a in RR, a>0.`

Enjoy Maths.!

Answer 2

`1`

Explanation:

We see that:

`lim_(xrarr0)1/x^2int_0^xtf(t)dt=lim_(xrarr0)1/x^2int_0^xtsqrt(4-t^2)dt`

Notice that attempting to evaluate the limit at `x=0` gives an indeterminate limit in the form `0/0`, where the denominator is `x^2` and the numerator is the integral.

Because we have this indeterminate form, we can use l'Hôpital's rule. Take the derivative of numerator and the denominator.

`=lim_(xrarr0)(1/(d/dxx^2))d/dxint_0^xtsqrt(4-t^2)dt`

The derivative of the integral can be found through the Second Fundamental Theorem of Calculus.

`=lim_(xrarr0)1/(2x)(xsqrt(4-x^2))=lim_(xrarr0)1/2sqrt(4-x^2)=sqrt4/2=1`