If x 0lime^(a-1)x+2 bx+(c-2)e^-xx x- _e(1+x)=2, then a^2+b^2+c^2 is equal to :

Question

If $\underset{x	o 0}{lim}\frac{{e}^{(a-1)x}+2\cos bx+(c-2){e}^{-x}}{x\cos x-{\log }_{e}(1+x)}=2$, then ${a}^{2}+{b}^{2}+{c}^{2}$ is equal to :

Accepted Answer

$\begin{gathered} \underset{x	o 0}{lim}[1+\frac{(a-1)}{1!}x+\frac{(a-1{)}^{2}}{2!}{x}^{2}+\ldots \ldots \ldots \infty] \ \frac{+2[1-\frac{(bx{)}^{2}}{2!}+\ldots \ldots \infty]+(c-2)[1-\frac{{x}^{1}}{1!}+\frac{{x}^{2}}{2!}-\ldots \ldots \infty]}{x(1-\frac{{x}^{2}}{2!}+\ldots \ldots \infty)-(x-\frac{{x}^{2}}{2}+\ldots \ldots \ldots \infty)} \ \Rightarrow \underset{x	o 0}{lim}\frac{(c+1)+(a-c+1)x+[\frac{(a-1{)}^{2}}{2}-{b}^{2}+\frac{(c-2)}{2}]{x}^{2}+\ldots \ldots \ldots \infty}{(\frac{{x}^{2}}{2}-\frac{{x}^{3}}{2}+\ldots \ldots \ldots \ldots \infty)}=2 \ \begin{gathered} c+1=0\Rightarrow c=-1 \ a-c+1=0\Rightarrow a+1+1=0\Rightarrow a=-2 \ \frac{(a-1{)}^{2}}{2}-{b}^{2}+\frac{(c-2)}{2}) \ (\frac{1}{2}) \ \Rightarrow 9-2{b}^{2}-3=2\Rightarrow 2{b}^{2}=4\Rightarrow {b}^{2}=2 \ 	herefore {a}^{2}+{b}^{2}+{c}^{2}=4+2+1=7 \end{gathered} \end{gathered}$

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