Mathematics - Limits Question with Solution | TestHub

${\lim }_{x\to 0}\frac{\frac{\alpha }{2}{\int }_0^x\frac{1}{1-t^2}dt+\beta x\cos x}{x^3}$ $\begin{array}{rl}& =\underset{x\to 0}{\lim }\frac{\frac{\alpha }{2}\left(\frac{1}{1-x^2}\right)+\beta \cos x-\beta x\sin x}{3x^2}\\ & =\frac{\frac{\alpha }{2}{\left(1-x^2\right)}^{-1}+\beta \left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\dots \right)-\beta x\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\dots \right)}{3x^2}\\ & =\frac{\frac{\alpha }{2}\left(1+x^2+x^4\dots \right)+\beta \left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\dots \right)-\beta \left(x^2-\frac{x^4}{3!}\dots \right)}{3x^2}\end{array}$ $\begin{array}{rl}& =\frac{\left(\frac{\alpha }{2}+\beta \right)+{\mathrm{x}}^2\left(\frac{\alpha }{2}-\frac{\beta }{2}-\beta \right)+{\mathrm{x}}^4()\dots }{3{\mathrm{x}}^2}=2(\text{ Given })\\ & \therefore \frac{\alpha }{2}+\beta =0\text{ and }\frac{\alpha -3\beta }{6}=2\\ & \Rightarrow \alpha =-2\beta \text{ and }\alpha =12+3\beta \\ & \Rightarrow \beta =-\frac{12}{5}\text{ and }\alpha =\frac{24}{5}\end{array}$ $\therefore \alpha +\beta =\frac{12}{5}=2.40$