lim-x-0-sin-x-x-1-how-is-this-so-

Question Number 4773 by madscientist last updated on 08/Mar/16

{l i m}_{x \to 0} \frac{s i n (x)}{x} = 1

h o w i s t h i s s o ?

Commented by Yozzii last updated on 13/Mar/16

F o r x n e a r z e r o, t h e f u n c t i o n s i n x

i s v e r y w e l l a p p r o x i m a t e d b y t h e

M a c l u a r i n e x p a n s i o n

x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \frac{x^{9}}{9!} + \dots = \underset{r = 0}{\sum^{\infty}} \frac{{(- 1)}^{r} x^{2 r + 1}}{(2 r + 1)!} .

∴ \frac{s i n x}{x} = \frac{1}{x} (\underset{r = 0}{\sum^{\infty}} \frac{{(- 1)}^{r} x^{2 r + 1}}{(2 r + 1)!}) = \underset{r = 0}{\sum^{\infty}} \frac{{(- 1)}^{r} x^{2 r}}{(2 r + 1)!}

o r \frac{s i n x}{x} = 1 + \underset{r = 1}{\sum^{\infty}} \frac{{(- 1)}^{r} x^{2 r}}{(2 r + 1)!} f o r x n e a r z e r o .

∴ \lim_{x \to 0} \frac{s i n x}{x} = \lim_{x \to 0} (1 + \underset{r = 1}{\sum^{\infty}} \frac{{(- 1)}^{r} x^{2 r}}{(2 r + 1)!})

= 1 + \underset{r = 1}{\sum^{\infty}} \frac{{(- 1)}^{r} 0^{2 r}}{(2 r + 1)!}

= 1 + 0 + 0 + 0 + \dots

\lim_{x \to 0} \frac{s i n x}{x} = 1 .

Answered by malwaan last updated on 08/Mar/16

{l i m}_{x \to 0} \frac{s i n x}{x} = {l i m}_{x \to 0} \frac{c o s x}{1} = c o s (0) = 1

Commented by Dnilka228 last updated on 10/Mar/16

\lim_{a} \frac{\sin a}{a} = \lim_{a} \frac{\cos a}{1} = \cos (0) = 2