lim-x-sin-x-1-x-sin-x-

Question Number 90077 by manuel__ last updated on 21/Apr/20

{l i m}_{x \to \infty} (\sin (x + \frac{1}{x}) - s i n (x)) = ?

Commented by jagoll last updated on 21/Apr/20

\sin (x + \frac{1}{x}) - \sin (x) =

2 \cos (\frac{2 x + \frac{1}{x}}{2}) \sin (\frac{1}{2 x}) =

2 \cos (x + \frac{1}{2 x}) \sin (\frac{1}{2 x})

[let \frac{1}{2 x} = t, t \to 0]

\lim_{t \to 0} 2 \cos (t + \frac{1}{2 t}) \sin t = 0

Commented by mathmax by abdo last updated on 21/Apr/20

l e t f (x) = s i n (x + \frac{1}{x}) - s i n x f i s c o n t i n u e \Rightarrow

{l i m}_{x \to + \infty} f (x) = {l i m}_{x \to + \infty} (s i n x - s i n x) = 0

Commented by mathmax by abdo last updated on 21/Apr/20

a n o t h e r w a y f (x) = s i n x c o s (\frac{1}{x}) + c o s x s i n (\frac{1}{x}) - s i n x

\sim s i n x (1 - \frac{1}{2 x^{2}}) + c o s x \times \frac{1}{x} - s i n x (x \to \infty) \Rightarrow

f (x) \sim - \frac{s i n x}{2 x^{2}} + \frac{c o s x}{x} w e h a v e ∣ \frac{- s i n x}{2 x^{2}} ∣⩽ \frac{1}{2 x^{2}} \to 0 a l s o

∣ \frac{c o s x}{x} ∣⩽ \frac{1}{∣ x ∣} \to 0 \Rightarrow {l i m}_{x \to \infty} f (x) = 0