How to derive this equation? Find a.

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Answer 1 · 2018-04-25T09:57:10

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

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#.

Answer 2 · 2018-04-27T18:28:17

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

#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))#

#= 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#