If a function f is defined by f(x) = frac{cot | vedclass

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(B) We need to evaluate $\lim_{x ightarrow \pi/4} \frac{\cot^3 x - 	an x}{\cos(x + \pi/4)}$.Let $t = 	an x$. As $x ightarrow \pi/4$,$t ightarr

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$8$

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(B) We need to evaluate $\lim_{x ightarrow \pi/4} \frac{\cot^3 x - 	an x}{\cos(x + \pi/4)}$.Let $t = 	an x$. As $x ightarrow \pi/4$,$t ightarrow 1$.The numerator is $\frac{1}{t^3} - t = \frac{1 - t^4}{t^3} = \frac{(1 - t^2)(1 + t^2)}{t^3} = \frac{(1 - t)(1 + t)(1 + t^2)}{t^3}$.The denominator is $\cos(x + \pi/4) = \cos x \cos(\pi/4) - \sin x \sin(\pi/4) = \frac{1}{\sqrt{2}}(\cos x - \sin x) = \frac{\cos x}{\sqrt{2}}(1 - 	an x) = \frac{\cos x}{\sqrt{2}}(1 - t)$.Substituting these into the limit:$\lim_{t ightarrow 1} \frac{(1 - t)(1 + t)(1 + t^2)}{t^3} \cdot \frac{\sqrt{2}}{\cos x(1 - t)} = \lim_{t ightarrow 1} \frac{\sqrt{2}(1 + t)(1 + t^2)}{t^3 \cos x}$.As $t ightarrow 1$,$x ightarrow \pi/4$,so $\cos x ightarrow 1/\sqrt{2}$.$= \frac{\sqrt{2}(1 + 1)(1 + 1^2)}{1^3 \cdot (1/\sqrt{2})} = \frac{\sqrt{2} \cdot 2 \cdot 2}{1/\sqrt{2}} = 4 \cdot 2 = 8$.

If a function $f$ is defined by $f(x) = \frac{\cot^3 x - \tan x}{\cos(x + \pi/4)}$ for $x \neq \pi/4$,then $\lim_{x \rightarrow \pi/4} f(x) = $

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