Lim x -2sin x-sinx xsin x-1-sin x-ln s i n x is equal to

The correct option is B 2Let sinx=t Then t→ 1limt t^2/1 t+ln t=t→ 1lim1 t^t(1+ln t)/0 1+1/t (by L'Hospital's rule) =t→ 1lim0 { t^t ( 1/t )+t^t(1+ln t)^2}/ ...

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Standard XII

Mathematics

Property 5

lim x →π/2sin...

Question

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

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Solution

The correct option is B 2
$L e t s i n x = t T h e n l i m t \to 1 \frac{t - t^{2}}{1 - t + l n t} = l i m t \to 1 \frac{1 - t^{t} (1 + l n t)}{0 - 1 + \frac{1}{t}} (b y L^{'} H o s p i t a l^{'} s r u l e) = l i m t \to 1 \frac{0 - {t^{t} (\frac{1}{t}) + t^{t} (1 + l n t)^{2}}}{0 - \frac{1}{t^{2}}} = 2$

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limx→π2sinx−(sinx)sinx1−sinx+lnsinx is equal to

The correct option is B 2Let sinx=tThen limt→1t−t21−t+lnt=limt→11−tt(1+lnt)0−1+1t (by L′Hospital′s rule)=limt→10−{tt(1t)+tt(1+lnt)2}0−1t2=2

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