Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 56202 by naka3546 last updated on 12/Mar/19

$$\underset{x\to 0}{\lim }\frac{x^2{\tan }^{-1}\left(x\right)-3\underset{0}{\int }\stackrel{x}{}\sin \left(t^2\right)dt}{x^5}=?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Mar/19

$$g\left(x\right)={\int }_0^{x}sin\left(t^2\right)dt$$$$\frac{dg}{dx}={\int }_0^x\frac{\partial }{\partial x}\left({sint}^2\right)dt+{sinx}^2\times \frac{dx}{dx}-sin{0}^2\times \frac{d\left(0\right)}{dx}$$$$={sinx}^2$$$$\underset{x\to 0}{\lim }\frac{x^2\times \frac{1}{1+x^2}+2{xtan}^{-1}\left(x\right)-3{sinx}^2}{5x^4}\left(\frac{0}{0}\right)form$$$$\underset{x\to 0}{\lim }\frac{\frac{(1+x^2)2x-x^2\left(2x\right)}{{(1+x^2)}^2}+\frac{2x}{1+x^2}+2{tan}^{-1}\left(x\right)-6{xcosx}^2}{20x^3}$$$$\underset{x\to 0}{\lim }\frac{\frac{2x}{{(1+x^2)}^2}+\frac{2x}{1+x^2}+2{tan}^{-1}\left(x\right)-6{xcosx}^2}{20x^3}\left(\frac{0}{0}\right)$$$$\underset{x\to 0}{\lim }\frac{2x+2x+2x^3+2{tan}^{-1}\left(x\right)\times {(1+x^2)}^2-6x{(1+x^2)}^2{cosx}^2}{20x^3{(1+x^2)}^2}$$$$wait...complicated...stilltodifferntiate...forLH$$$$rule...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com