X-1lim1-x tan- - x-2 - is equal to -

The correct option is A 2π x→ 1lim(1 x) tan[ π x/2 ]= h→ 0lim(1 1+h) tan[ π/2 π h/2 ] =h→ 0lim h π/2h= h→ 0limh/tanπ h/2=2/π ...

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Solving Inequalities

x→1lim1-x tan...

Question

$lim x \to 1 (1 - x) tan (\begin{matrix} \frac{π x}{2} \end{matrix})$ is equal to -

\frac{2}{π}

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\frac{π}{2}

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π

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3 π

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A = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦, B = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

A - B

lim x \to π / 2 \frac{(1 - tan \frac{x}{2})}{(1 + tan \frac{x}{2})} \frac{(1 - sin x)}{(π - 2 x)^{3}}

\int_{\frac{- 3 π}{2}}^{- \frac{π}{2}} [(x + π)^{3} + c o s^{2} (x + π)] d x

\int_{\frac{- 3 π}{2}}^{- \frac{π}{2}} [(x + π)^{3} + c o s^{2} (x + π)] d x

\int_{- 3 π / 2}^{- π / 2} [(x + π)^{3} + c o s^{2} (x + 3 π)]

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