If alpha = lim_{x rightarrow pi/4} frac{tan^{ | vedclass

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(D) First,calculate $\alpha = \lim_{x ightarrow \pi/4} \frac{	an^{3} x - 	an x}{\cos(x + \pi/4)}$. This is a $\frac{0}{0}$ form.Applying $L'H\hat{

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$(1, 3)$

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(D) First,calculate $\alpha = \lim_{x ightarrow \pi/4} \frac{	an^{3} x - 	an x}{\cos(x + \pi/4)}$. This is a $\frac{0}{0}$ form.Applying $L'H\hat{o}pital's$ rule:$\alpha = \lim_{x ightarrow \pi/4} \frac{3 	an^{2} x \sec^{2} x - \sec^{2} x}{-\sin(x + \pi/4)} = \frac{3(1)^{2}(2) - 2}{-\sin(\pi/2)} = \frac{6 - 2}{-1} = -4$.Next,calculate $\beta = \lim_{x ightarrow 0} (\cos x)^{\cot x}$. This is a $1^{\infty}$ form.$\beta = e^{\lim_{x ightarrow 0} \cot x (\cos x - 1)} = e^{\lim_{x ightarrow 0} \frac{\cos x - 1}{	an x}} = e^{\lim_{x ightarrow 0} \frac{-\sin x}{\sec^{2} x}} = e^{0} = 1$.Since $\alpha = -4$ and $\beta = 1$ are roots of $ax^{2} + bx - 4 = 0$,the sum of roots $\alpha + \beta = -b/a \Rightarrow -4 + 1 = -3 = -b/a \Rightarrow b = 3a$.The product of roots $\alpha \beta = -4/a \Rightarrow (-4)(1) = -4/a \Rightarrow a = 1$.Substituting $a = 1$ into $b = 3a$,we get $b = 3$.Thus,the ordered pair $(a, b)$ is $(1, 3)$.

If $\alpha = \lim_{x \rightarrow \pi/4} \frac{\tan^{3} x - \tan x}{\cos(x + \pi/4)}$ and $\beta = \lim_{x \rightarrow 0} (\cos x)^{\cot x}$ are the roots of the equation $ax^{2} + bx - 4 = 0$,then the ordered pair $(a, b)$ is:

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