If undersetx→ 0 lim α e^x+β e^-x+γ sinx x sin^2x= 2 3, where α ,β ,γ ∈ R, then which of the following is NOT correct?

Given,

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\alpha {e}^{x}+\beta {e}^{-x}+\gamma \mathrm{sin}x}{x{\mathrm{sin}}^{2}x}=\frac{2}{3}$$x→0lim​xsin2xαex+βe−x+γsinx​=32​

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{\alpha {e}^{x}+\beta {e}^{-x}+\gamma \mathrm{sin}x}{{x}^{3}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}}=\frac{2}{3}$$⇒x→0lim​x3x2sin2x​αex+βe−x+γsinx​=32​

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{\alpha {e}^{x}+\beta {e}^{-x}+\gamma \mathrm{sin}x}{{x}^{3}}=\frac{2}{3}$$⇒x→0lim​x3αex+βe−x+γsinx​=32​

Now using expansion of given function in terms of $x$x we get,

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\alpha (1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+\ldots )+\beta (1-x+\frac{{x}^{2}}{2!}-\frac{{x}^{3}}{3!}+\ldots )+\gamma (x-\frac{{x}^{3}}{3!}+\ldots )}{{x}^{3}}$$=x→0lim​x3α(1+x+2!x2​+3!x3​+…)+β(1−x+2!x2​−3!x3​+…)+γ(x−3!x3​+…)​

Now for limit to exist constant terms should be zero

$$\Rightarrow a+\beta =0....(1)$$⇒a+β=0....(1)

Also coefficient of $x$x should be zero

$$\Rightarrow \alpha -\beta +\gamma =0.......(2)$$⇒α−β+γ=0.......(2)

Also coefficient of ${x}^{2}$x2 should be zero

$$\Rightarrow \frac{\alpha }{2}+\frac{\beta }{2}=0.......(3)$$⇒2α​+2β​=0.......(3)

So limit becomes, $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{x}^{3}(\frac{\alpha }{3!}-\frac{\beta }{3!}-\frac{\gamma }{3!})+{x}^{4}(\frac{\alpha }{3!}-\frac{\beta }{3!}-\frac{\gamma }{3!})}{{x}^{3}}=\frac{2}{3}$x→0lim​x3x3(3!α​−3!β​−3!γ​)+x4(3!α​−3!β​−3!γ​)​=32​

Now putting the value of limit we get,

$$\frac{\alpha }{6}-\frac{\beta }{6}-\frac{\gamma }{6}=\frac{2}{3}........(4)$$6α​−6β​−6γ​=32​........(4)

Now on solving equation 1,2,3&4 we get,

$$\Rightarrow \alpha =1,\beta =-1,\gamma =-2$$⇒α=1,β=−1,γ=−2

Now putting these in all option we get, $\alpha {\beta }^{2}+\beta {\gamma }^{2}+\gamma {\alpha }^{2}+3=0$αβ2+βγ2+γα2+3=0