1) lim_(x→0) (x/(e^(1/x) +1)) = ? 2) For xεR, f(x)=∣ln2−sin x∣ and g(x)=f(f(x)), then prove that g′(0)=cos (ln2).


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Question Number 57442 by rahul 19 last updated on 04/Apr/19

$1) \lim_{x \to 0} \frac{x}{e^{\frac{1}{x}} + 1} = ?$ $2) F o r x ϵ R, f (x) =∣ l n 2 - \sin x ∣ a n d$ $g (x) = f (f (x)), t h e n p r o v e t h a t$ $g^{'} (0) = \cos (l n 2) .$
Commented by rahul 19 last updated on 04/Apr/19

$i n t h e n e i g h b o u r h o o d o f 0,$ $w h i c h i s g r e a t e r :$ $l n 2 o r \sin (l n 2 - \sin x) ?$ $(c a l u l a t o r n o t a l l o w e d) .$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

$w h i o u t u s i n g g r a p h a p p v e r y d i f f i c u l t t o r e a c h$ $t h e g o a l . . .$
Commented by rahul 19 last updated on 04/Apr/19

$o k s i r .$
Commented by kaivan.ahmadi last updated on 04/Apr/19

${l i m}_{x \to 0^{+}} \frac{x}{e^{\frac{1}{x}} + 1} = \frac{0^{+}}{e^{+ \infty} + 1} = \frac{0^{+}}{+ \infty} = 0$ $a n d$ ${l i m}_{x \to 0^{-}} \frac{x}{e^{\frac{1}{x}} + 1} = \frac{0^{-}}{e^{- \infty} + 1} = \frac{0^{-}}{0 + 1} = 0$ $s o w e h a v e$ ${l i m}_{x \to 0} \frac{x}{e^{\frac{1}{x}} + 1} = 0$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

	$1) \underset{x \to 0}{li}$ $l e t t = \frac{1}{x} w h e n x \to 0 t \to \infty$ $l i \underset{t \to \infty}{m} \frac{1}{t} \times \frac{1}{e^{t} + 1}$ $w h e n t \to \infty \frac{1}{t} \to 0$ $w h e n t \to \infty \frac{1}{e^{t} + 1} \to 0$ $s o l i \underset{t \to \infty}{m} \frac{1}{t} \times \frac{1}{e^{t} + 1} = 0$
	Commented by rahul 19 last updated on 04/Apr/19

	$t h a n k s s i r!$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

	$s i n x v a l u e l i e s b e t w e e n \pm 1$ $w e k n o w l n 2 = 0.6931$ $f (x) = l n 2 - s i n x w h e n l n 2 > s i n x > 0$ $= 0 w h e n s i n x = l n 2$ $= - (l n 2 - s i n x) w h e n 1 > s i n x > l n 2$ $n o w l o o k w h e n s i n x < 0$ $t h e n f (x) =∣ l n 2 - s i n x ∣= (l n 2 - s i n x)$ $h e r e w e r e s t r i c t t o c o n s i d e r + v e v a l u e o f$ $s i n x$ $A) v a l u e o f \frac{d g}{d x} w h e n s i n x < l n 2$ $g (x) = f [f (x)]$ $g (x) = f (l n 2 - s i n x) w h e n [l n 2 > s i n x > 0]$ $= l n 2 - s i n (l n 2 - s i n x)$ $\frac{d g (x)}{d x} = 0 - c o s (l n 2 - s i n x) \times (0 - c o s x)$ ${(\frac{d g}{d x})}_{x = 0} = - c o s (l n 2 - 0) \times - 1 = c o s l n 2$ $B) v a l u e o f \frac{d g}{d x} w h e n s i n x = l n 2$ $= f [f (x)] w h e n s i n x = l n 2$ $= f [0]$ $= l n 2 - s i n 0$ $= l n 2$ $s o \frac{d g}{d x} = 0 a t s i n x = l n 2$ $C) v a l u e o f \frac{d g}{d x} w h e n s i n x > l n 2$ $g (x) = f (- l n 2 + s i n x)$ $= l n 2 - s i n (- l n 2 + s i n x)$ $\frac{d g}{d x} = 0 - c o s (- l n 2 + s i n x) \times (0 + c o s x)$ ${(\frac{d g}{d x})}_{x = 0} = - c o s (- l n 2) = - c o s l n 2$
	Commented by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

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