x 0lim3+ x+ x+ _e(1-x)3 ^2x=(1)/(3)2 -

Question

$\underset{x	o 0}{lim}\frac{3+\alpha \sin x+\beta \cos x+{\log }_{e}(1-x)}{3{	an }^{2}x}=\frac{1}{3}2\alpha -\beta $

Accepted Answer

$\underset{x	o 0}{lim}\frac{3+\alpha \sin x+\beta \cos x+{\log }_{e}(1-x)}{3{	an }^{2}x}=\frac{1}{3}$

$\Rightarrow \underset{x	o 0}{lim}\frac{3+\alpha [x-\frac{{x}^{3}}{3!}+\ldots .]+\beta [1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}\ldots .]+(-x-\frac{{x}^{2}}{2}-\frac{{x}^{3}}{3}\ldots )}{3{	an }^{2}x}=\frac{1}{3}$

$\Rightarrow \underset{x	o 0}{lim}\frac{(3+\beta )+(\alpha -1)x+(-\frac{1}{2}-\frac{\beta }{2}){x}^{2}+\ldots .}{3{x}^{2}}	imes \frac{{x}^{2}}{{	an }^{2}x}=\frac{1}{3}$

$\begin{gathered} \Rightarrow \beta +3=0,\alpha -1=0\frac{-\frac{1}{2}-\frac{\beta }{2}}{3}=\frac{1}{3} \ \Rightarrow \beta =-3,\alpha =1 \ \Rightarrow 2\alpha -\beta =2+3=5 \end{gathered}$

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