Limx -2sin x - sin xsin x-1-sin x-1n sin x is equal to

Let sin x be = t then = lim_t→ 1t t^t/1 t+In t (by L'Hospital's rule) = lim_t→ 11 t^t(1+In t)/0 1+1/t = lim_t→ 10 {t^t(1/t)+t^t(1+In t)^2/0 1/t^2 = 2 ...

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Standard XII

Mathematics

L'Hospital Rule to Remove Indeterminate Form

limx →π/2sin ...

Question

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

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Solution

The correct option is B
2

Let sin x be = t

then = $lim t \to 1 \frac{t - t^{t}}{1 - t + I n t}$
(by L'Hospital's rule)
= $lim t \to 1 \frac{1 - t^{t} (1 + I n t)}{0 - 1 + \frac{1}{t}}$
= $lim t \to 1 \frac{0 - {t^{t} (\frac{1}{t}) + t^{t} (1 + I n t)^{2}}{0 - \frac{1}{t^{2}}}$ = 2

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limx→π2sin x−(sin x)sin x1−sin x+1n sin x is equal to

The correct option is B 2 Let sin x be = t then = limt→1t−tt1−t+In t (by L'Hospital's rule) = limt→11−tt(1+In t)0−1+1t = limt→10−{tt(1t)+tt(1+In t)20−1t2 = 2

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

The correct option is B
2

Let sin x be = t

then = $lim t \to 1 \frac{t - t^{t}}{1 - t + I n t}$
(by L'Hospital's rule)
= $lim t \to 1 \frac{1 - t^{t} (1 + I n t)}{0 - 1 + \frac{1}{t}}$
= $lim t \to 1 \frac{0 - {t^{t} (\frac{1}{t}) + t^{t} (1 + I n t)^{2}}{0 - \frac{1}{t^{2}}}$ = 2