For , , R, if x 0limx^2 x+( -1)e^x^2 2x- x=3, then + - is equal to :

Question

For $\alpha ,\beta ,\gamma \in R$, if $\underset{x	o 0}{lim}\frac{{x}^{2}\sin \alpha x+(\gamma -1){e}^{{x}^{2}}}{\sin 2x-\beta x}=3$, then $\beta +\gamma -\alpha $ is equal to :

Accepted Answer

$\begin{gathered} x	o 0 \ \sin 2x-\beta x	o 0 \ \Rightarrow \, \frac{0}{0} \ \Rightarrow \, (\gamma -1){e}^{0}+0\sin (\alpha x)	o 0 \ \Rightarrow \, (\gamma -1)=0 \ \Rightarrow \, \gamma =1 \ \Rightarrow \, \underset{x	o 0}{lim}\frac{{x}^{2}\sin (\alpha x)}{(\sin 2x-\beta x)}=3 \end{gathered}$

$\Rightarrow \underset{x	o 0}{lim}\frac{{x}^{2}[\alpha x-\frac{(\alpha x{)}^{3}}{3!}+\frac{(\alpha x{)}^{5}}{5!}-\cdots ]}{[(2x)-\frac{(2x{)}^{3}}{3!}+\frac{(2x{)}^{5}}{5!}]-\beta x}$

$\Rightarrow \underset{x	o 0}{lim}\frac{\alpha {x}^{3}-\frac{{\alpha }^{3}{x}^{5}}{3!}+\frac{{\alpha }^{5}{x}^{7}}{5!}-\cdots }{x(2-\beta )-\frac{8{x}^{3}}{6}+\frac{{2}^{5}\cdot {x}^{5}}{5!}-\cdots }=3$

$\begin{gathered} \Rightarrow 2-\beta =0\frac{\alpha }{\frac{-8}{6}}=3 \ \Rightarrow \beta =2 \ \alpha =3(-\frac{8}{6})=-4 \ \Rightarrow \gamma =1,\beta =2,\alpha =-4 \ \Rightarrow \beta +\gamma -\alpha =7 \end{gathered}$

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