Mathematics - Limits Question with Solution | TestHub

Given,

$\underset{\mathrm{x}\to 0}{\lim }\frac{3+\mathrm{\alpha sinx}+\mathrm{\beta cosx}+{\log }_{\mathrm{e}}(1-\mathrm{x})}{3{\tan }^2\mathrm{x}}=\frac{1}{3}$

$\Rightarrow \underset{\mathrm{x}\to 0}{\lim }\frac{3+\mathrm{\alpha }\left[\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\dots \right]+\mathrm{\beta }\left[1-\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}\dots \right]+\left(-\mathrm{x}-\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^3}{3}\dots \right)}{3{\tan }^2\mathrm{x}}=\frac{1}{3}$

$\Rightarrow \underset{\mathrm{x}\to 0}{\lim }\frac{(3+\mathrm{\beta })+(\mathrm{\alpha }-1)\mathrm{x}+\left(-\frac{1}{2}-\frac{\mathrm{\beta }}{2}\right){\mathrm{x}}^2+\dots .}{3{\mathrm{x}}^2}\times \frac{{\mathrm{x}}^2}{{\tan }^2\mathrm{x}}=\frac{1}{3}$

The coefficient of all the powers of $\mathrm{x}$ except ${\mathrm{x}}^2$ is $0$.

$\Rightarrow \mathrm{\beta }+3=0,\mathrm{\alpha }-1=0$ and $\frac{-\frac{1}{2}-\frac{\mathrm{\beta }}{2}}{3}=\frac{1}{3}$

$\Rightarrow \mathrm{\beta }=-3,\mathrm{\alpha }=1$

$\Rightarrow 2\mathrm{\alpha }-\mathrm{\beta }=2+3=5$