lim _{x} {rightarrow 0} frac{tan left(left[-p | vedclass

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(B) Given the limit: $\lim _{x ightarrow 0} \frac{	an ([- \pi^2] x^2) - x^2 	an ([- \pi^2])}{\sin ^2 x}$.Since $\pi^2 \approx 9.87$,the greatest i

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$	an 10-10$

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(B) Given the limit: $\lim _{x ightarrow 0} \frac{	an ([- \pi^2] x^2) - x^2 	an ([- \pi^2])}{\sin ^2 x}$.Since $\pi^2 \approx 9.87$,the greatest integer value $[-\pi^2] = [-9.87] = -10$.Substituting this value,the expression becomes: $\lim _{x ightarrow 0} \frac{	an (-10 x^2) - x^2 	an (-10)}{\sin ^2 x}$.Using $	an (- 	heta) = - 	an 	heta$,we get: $\lim _{x ightarrow 0} \frac{-	an (10 x^2) + x^2 	an 10}{\sin ^2 x}$.Using the limit $\lim _{x ightarrow 0} \frac{	an (kx^2)}{x^2} = k$ and $\lim _{x ightarrow 0} \frac{\sin ^2 x}{x^2} = 1$,we divide the numerator and denominator by $x^2$:$= \lim _{x ightarrow 0} \frac{-\frac{	an (10 x^2)}{x^2} + 	an 10}{\frac{\sin ^2 x}{x^2}}$.$= \frac{-10 + 	an 10}{1} = 	an 10 - 10$.

$\lim _{x}$ ${\rightarrow 0} \frac{\tan \left(\left[-\pi^2\right] x^2\right)-x^2 \tan \left(\left[-\pi^2\right]\right)}{\sin ^2 x}$ equals

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