if lim_(x→0) ((ae^x −bsin x+ce^(−x) )/(xsin x)) = 2 what is a+b+c ?


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Question Number 80983 by jagoll last updated on 08/Feb/20

$i f \lim_{x \to 0} \frac{{a e}^{x} - b \sin x + {c e}^{- x}}{x \sin x} = 2$ $w h a t i s a + b + c ?$
Commented by jagoll last updated on 09/Feb/20

$i {c a n}^{'} t s o l v e d t h i s e q u a t i o n .$ $w h a t {i t}^{'} s w r o n g ?$
Commented by jagoll last updated on 09/Feb/20

$s t e p (1) l i m i t m u s t b e \frac{0}{0}$ $\Rightarrow a + c = 0 \Rightarrow c = - a$ $s t e p (2) \lim_{x \to 0} \frac{{a e}^{x} - b \sin x - {a e}^{- x}}{x \sin x} = 2$ $L \hat{} h o p i t a l r u l e$ $\lim_{x \to 0} \frac{{a e}^{x} - b \cos x + {a e}^{- x}}{\sin x + x \cos x} = 2$ $s i n c e d e n u m e r a t o r = 0,$ $\Rightarrow 2 a - b = 0 \Rightarrow b = 2 a$ $L \hat{} h o p i t a l a g a i n$ $\lim_{x \to 0} \frac{{a e}^{x} + b \sin x - {a e}^{- x}}{2 \cos x - x \sin x} = 2$ $\Rightarrow \frac{0}{2} = 2 \Rightarrow ?$
Commented by abdomathmax last updated on 09/Feb/20

$\Rightarrow {l i m}_{x \to 0} \frac{{a e}^{x} - b s i n x + c e^{- x}}{x s i n x} - 2 = 0 \Rightarrow$ ${l i m}_{x \to 0} \frac{a e^{x} - b s i n x + {c e}^{- x} - 2 x s i n x}{x s i n x} = 0 \Rightarrow$ ${l i m}_{x \to 0} u (x) == {l i m}_{x \to 0} u^{^{'}} (x) = 0 a n d$ ${l i m}_{x \to 0} v (x) = {l i m}_{x \to 0} v^{^{'}} (x) = 0 w i t h$ $u (x) = {a e}^{x} - b s i n x + {c e}^{- x} - 2 x s i n x \Rightarrow$ $u (0) = 0 \Rightarrow a + c = 0 \Rightarrow c = - a$ $u^{^{'}} (x) = {a e}^{x} - b c o s x - {c e}^{- x} - 2 s i n x - 2 x c o s x$ $u^{^{'}} (0) = 0 \Rightarrow a - b - c = 0 \Rightarrow 2 a - b = 0 \Rightarrow b = 2 a$ $\Rightarrow a + b + c = 2 (b + c) = 2 (2 a - a) = 2 a$ $i f w e g e t a = 1 \Rightarrow a + b + c = 2$
Commented by john santu last updated on 09/Feb/20

$y o u a r e r i g h t s i r$
Commented by jagoll last updated on 09/Feb/20

$h o w t o g e t a = 1 ?$
Commented by jagoll last updated on 09/Feb/20

$i f w e g e t a = 1, i t m e a n f o r a = 6$ $a + b + c = 12 . t h a t m e a n s t h e$ $s o l u t i o n f r o m a + b + c i s n o t$ $o n e .$
Commented by mr W last updated on 09/Feb/20

$h a v e y o u c h e c k e d s i r ?$ $i {d o n}^{'} t t h i n k t h e a n s w e r i s c o r r e c t .$ $e x a m p l e : a = 1, b = 2, c = - 1 .$ $\lim_{x \to 0} \frac{e^{x} - 2 \sin x - e^{- x}}{x \sin x}$ $= \lim_{x \to 0} \frac{e^{x} - 2 \cos x + e^{- x}}{\sin x + x \cos x}$ $= \lim_{x \to 0} \frac{e^{x} + 2 \sin x - e^{- x}}{2 \cos x - x \sin x}$ $= \frac{0}{2} = 0 \neq 2$ $m y a n s w e r i s :$ $t h e r e e x i s t n o v a l u e s f o r a, b, c s u c h$ $t h a t \lim_{x \to 0} \frac{{a e}^{x} - b \sin x + {c e}^{- x}}{x \sin x} = 2 .$
Commented by jagoll last updated on 09/Feb/20

$y e s s i r . i t h i n k i t q u e s t i o n n o t r i g h t$
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