The value of limx→1512π 1-xtanπx2 must be

Put x=1+h Then limh→0-htanπ21+h=limh→0-h-cotπh2=limh→0htanπh2=2πlimh→0πh2tanπh2=2π×1=2π ∴limx→1512π 1-xtanπx2=512π×2π=1024

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The value of $\lim_{x \to 1} 512 π (1 - x) \tan (\frac{πx}{2})$ must be

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answer is 1024.

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Detailed Solution

Put $x = 1 + h$ Then $\lim_{h \to 0} (- h) \tan (\frac{π}{2} (1 + h)) = \lim_{h \to 0} (- h) (- \cot \frac{πh}{2})$ $= \lim_{h \to 0} \frac{h}{\tan (\frac{πh}{2})} = \frac{2}{π} \lim_{h \to 0} \frac{\frac{πh}{2}}{\tan (\frac{πh}{2})} = \frac{2}{π} \times 1 = \frac{2}{π} ∴ \lim_{x \to 1} 512 π (1 - x) \tan (\frac{πx}{2}) = 512 π \times \frac{2}{π} = 1024$