Solve the Following Questions
Question:

If $\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x \sin ^{2} x}=10, \alpha, \beta, \gamma \in \mathbf{R}$ then the value of $\alpha+\beta+\gamma$ is

Solution:

$\lim _{x \rightarrow 0} \frac{\alpha x\left(1+x+\frac{x^{2}}{2}\right)-\beta\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}\right)+\gamma x^{2}(1-x)}{x^{3}}$

$\lim _{x \rightarrow 0} \frac{x(\alpha-\beta)+x^{2}\left(\alpha+\frac{\beta}{2}+\gamma\right)+x^{3}\left(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma\right)}{x^{3}}=10$

For limit to exist

$\alpha-\beta=0, \alpha+\frac{\beta}{2}+\gamma=0$

$\frac{\alpha}{2}-\frac{\beta}{3}-\gamma=10 \ldots \ldots$ (i)

$\beta=\alpha, \gamma=-3 \frac{\alpha}{2}$

Put in (i)

$\frac{\alpha}{2}-\frac{\alpha}{3}+\frac{3 \alpha}{2}=10$

$\frac{\alpha}{6}+\frac{3 \alpha}{2}=10 \Rightarrow \frac{\alpha+9 \alpha}{6}=10$

$\rightarrow r=6$

$\alpha=6, \beta=6, \gamma=-9$

$\alpha+\beta+\gamma=3$

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