X-2lim - 1-tan - x-2 - -1-sin-x- - 1-tan - x-2 - -2x-3is1-8-0-1-32-

The correct option is D 1/32 Put x=pi/2 h as x→pi/2h→0 ∴Given limit=h→ 0lim1 tan ( π/4+h/2 )/1+tan ( π/4+h/2 )(1 cosh)/(2h)^3 =h→ 0lim tanh/22sin^2h/2/8h^3 = ...

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Byju's Answer

Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

x→π/2lim ( 1-...

Question

$l i m x \to \frac{π}{2} \frac{(1 - t a n (\frac{x}{2})) (1 - s i n x)}{(1 + t a n (\frac{x}{2})) (π - 2 x)^{3}} i s$

1/8

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Solution

The correct option is C
1/32

$Put x = \frac{p i}{2} - h a s x \to \frac{p i}{2} h \to 0 ∴ Given limit = l i m h \to 0 \frac{1 - t a n (\frac{π}{4} + \frac{h}{2})}{1 + t a n (\frac{π}{4} + \frac{h}{2})} \frac{(1 - c o s h)}{(2 h)^{3}} = l i m h \to 0 t a n \frac{h}{2} \frac{2 s i n^{2} \frac{h}{2}}{8 h^{3}} = l i m h \to 0 \frac{1}{4} . \frac{t a n \frac{h}{2}}{\frac{h}{2} \times 2} \times (\frac{s i n \frac{h}{2}}{\frac{h}{2}}) \times \frac{1}{4} l i m h \to 0 \frac{1}{32} (\frac{t a n \frac{h}{2}}{\frac{h}{2}}) {(\frac{s i n \frac{h}{2}}{\frac{h}{2}})}^{2} = \frac{1}{32}$

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Similar questions

Q. If

t a n x + t a n (x + \frac{π}{3}) + t a n (x + \frac{2 π}{3}) = 3,

then prove that

\frac{3 t a n x - t a n^{3} x}{1 - 3 t a n^{2} x} = 1

$l i m x \to \frac{π}{2} \frac{(1 - t a n (\frac{x}{2})) (1 - s i n x)}{(1 + t a n (\frac{x}{2})) (π - 2 x)^{3}} i s$

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

is equal to -

Q. Evaluate:

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

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

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limx→π2(1−tan(x2))(1−sin x)(1+tan(x2))(π−2x)3is

The correct option is C 1/32 Put x=pi2−h as x→pi2h→0∴Given limit=limh→01−tan(π4+h2)1+tan(π4+h2)(1−cosh)(2h)3=limh→0 tanh22sin2h28h3=limh→014.tanh2h2×2×(sinh2h2)×14limh→0132(tanh2h2)(sinh2h2)2=132

$l i m x \to \frac{π}{2} \frac{(1 - t a n (\frac{x}{2})) (1 - s i n x)}{(1 + t a n (\frac{x}{2})) (π - 2 x)^{3}} i s$