How to derive this equation? Find a.

limitlimit

2 Answers
Apr 25, 2018

I have requested Respected Steve M. Sir to check. Please wait.

AA a, b in RR, lim_(x to 0){int_0^xt^2/sqrt(a+2t^5)dt}/(bx-esinx)!=1/pi.

Explanation:

Suppose that, intt^2/sqrt(a+2t^5)dt=f(t)+c.

:." by the Definition of Integral, "f'(t)=t^2/sqrt(a+2t^5)......(star).

Further, using the Fundamental Theorem of Calculus, we have,

int_0^xt^2/sqrt(a+2t^5)dt=f(x)-f(0).............................(starstar).

Now, lim_(x to 0){int_0^xt^2/sqrt(a+2t^5)dt}/(bx-esinx),

=lim_(x to 0){f(x)-f(0)}/{x(b-e*sinx/x)}............[because, (starstar)],

=lim_(x to 0){(f(x)-f(0))/(x-0)}/{(b-e*sinx/x)},

={lim_(x to 0)(f(x)-f(0))/(x-0)}/{lim_(x to 0)(b-e*sinx/x)},

=(f'(0))/(b-e*1).

Here, by (star), f'(0)=0^2/sqrt(a+2*0^5)=0.

:. lim_(x to 0){int_0^xt^2/sqrt(a+2t^5)dt}/(bx-esinx)=0.

This means that, there do not exist a, b in RR, such that the

limit in question be 1/pi.

Apr 27, 2018

b = e, qquad a = ((2pi)/(e))^2

Explanation:

lim_(x to 0){int_0^x \ t^2/sqrt(a+2t^5)dt}/(bx-esinx)

= lim_(x to 0) {int_0^x \ t^2/sqrta - t^7/root(3)(a) + mathbb O(t^12) \ dt}/(bx-e(x - x^3/6 + mathbb O(x^5)))

= lim_(x to 0) { \ x^3/(3 sqrta) + mathbb O(x^8) } /(bx-ex + e x^3/6 + mathbb O(x^5))

  • Let b = e

= lim_(x to 0) { \ x^3/(3 sqrta) + mathbb O(x^8) } /( e x^3/6 + mathbb O(x^5)) = lim_(x to 0) { \ 1/(3 sqrta) } /( e/6 ) = 1/pi

implies a = ((2pi)/(e))^2