question lim_(x→0) ((sin(x+A)−sin(A−x))/(2x))


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Question Number 63473 by Rio Michael last updated on 04/Jul/19

$q u e s t i o n$ $\lim_{x \to 0} \frac{s i n (x + A) - s i n (A - x)}{2 x}$
Commented by mathmax by abdo last updated on 04/Jul/19

$l e t A (x) = \frac{s i n (x + A) - s i n (A - x)}{2 x} \Rightarrow$ $A (x) = \frac{s i n x c o s A + c o s x s i n A + s i n x c o s A - c o s x s i n A}{2 x}$ $= \frac{s i n x}{x} c o s A d u e t o {l i m}_{x \to 0} \frac{s i n x}{x} = 1 w e g e t {l i m}_{x \to 0} A (x) = c o s A .$ $l e t u s e h o s p i t a l t h e o r e m$ ${l i m}_{x \to 0} A (x) = {l i m}_{x \to 0} \frac{c o s (x + A) + c o s (x - A)}{2} = \frac{2 c o s A}{2} = c o s A .$
Commented by Rio Michael last updated on 04/Jul/19

$y e a h .$
Commented by Rio Michael last updated on 04/Jul/19

$i u s e d t h e i d e a o f s m a l l a n g l e s$ $f o r s m a l l a n g l e s s i n θ = θ$
Commented by mathmax by abdo last updated on 04/Jul/19

$y o u w i l l g e t t h e s a m e r e s u l t .$
Commented by Rio Michael last updated on 05/Jul/19

$b y s m a l l a n g l e s,$ $A (x) = \frac{s i n x c o s A + c o s x s i n A + s i n x c o s A - c o s x s i n A}{2 x}$ $= \frac{2 s i n x c o s A}{2 x}$ $\Rightarrow \underset{x \to 0}{l i m} \frac{2 s i n x c o s A}{2 x} f o r s m a l l a n g l e s s i n x = x$ $\Rightarrow \underset{x \to 0}{l i m} \frac{2 x c o s A}{2 x} = c o s A .$
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