The value of limx - 1 1-xtan - x2 is

Lim_x → 1 (1 x)tan (π x2 ), (0×∞ Form) =lim_x → 1(1 x) (π x2 ) Substitute x=1+h =lim_h → 0(1 1 h) (π (1+h)2 ) =lim_h → 0 h tan (π h2 ) =lim_h → 0π2htan (π h2 ) ...

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Standard XII

Mathematics

Differentiation under Integral Sign

The value of ...

Question

The value of $lim x \to 1 (1 - x) tan (\frac{π x}{2})$ is

π

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\frac{1}{π}

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\frac{2}{π}

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0

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Solution

The correct option is C $\frac{2}{π}$
$lim x \to 1 (1 - x) tan (\frac{π x}{2}), (0 \times \infty Form)$
$= lim x \to 1 \frac{(1 - x)}{cot (\frac{π x}{2})}$
Substitute $x = 1 + h$
$= lim h \to 0 \frac{(1 - 1 - h)}{cot (\frac{π (1 + h)}{2})}$
$= lim h \to 0 \frac{- h}{- tan (\frac{π h}{2})}$
$= lim h \to 0 \frac{\frac{π}{2} h}{tan (\frac{π h}{2})} \times \frac{1}{\frac{π}{2}}$
$= \frac{2}{π}$

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The value of limx→1(1−x)tan(πx2) is

The correct option is C 2πlimx→1(1−x)tan(πx2), (0×∞ Form) =limx→1(1−x)cot(πx2) Substitute x=1+h =limh→0(1−1−h)cot(π(1+h)2) =limh→0−h−tan(πh2) =limh→0π2htan(πh2)×1π2 =2π

The value of $lim x \to 1 (1 - x) tan (\frac{π x}{2})$ is