For alpha, beta, gamma in R,if lim _{x righta | vedclass

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(A) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2 \sin(\alpha x) + (\gamma-1) e^{x^2}}{\sin(2x) - \beta x} = 3$. For the limit to exist and be f

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$7$

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(A) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2 \sin(\alpha x) + (\gamma-1) e^{x^2}}{\sin(2x) - \beta x} = 3$. For the limit to exist and be finite,the numerator must approach $0$ as $x \rightarrow 0$. Since $\lim _{x \rightarrow 0} (x^2 \sin(\alpha x) + (\gamma-1) e^{x^2}) = 0 + (\gamma-1)(1) = \gamma-1$,we must have $\gamma-1 = 0$,so $\gamma = 1$. Now the expression becomes $\lim _{x \rightarrow 0} \frac{x^2(\alpha x + O(x^3)) + 0(e^{x^2}-1)}{(2x - \frac{8x^3}{6} + O(x^5)) - \beta x} = \lim _{x \rightarrow 0} \frac{\alpha x^3}{(2-\beta)x - \frac{4}{3}x^3} = 3$. For the limit to be non-zero,the denominator must have the same power of $x$ as the numerator. Thus,the coefficient of $x$ must be $0$,so $2-\beta = 0$,which gives $\beta = 2$. Now the limit is $\lim _{x \rightarrow 0} \frac{\alpha x^3}{-\frac{4}{3}x^3} = -\frac{3\alpha}{4} = 3$. Solving for $\alpha$,we get $\alpha = -4$. Finally,$\beta + \gamma - \alpha = 2 + 1 - (-4) = 7$.

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

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