lim-x-0-x-2-sin-x-4-x-

Question Number 102987 by bobhans last updated on 12/Jul/20

\lim_{x \to 0} \frac{x^{2} \sin (x^{- 4})}{x} ?

Answered by bemath last updated on 12/Jul/20

\lim_{x \to 0} \frac{2 x \sin (x^{- 4}) - 4 x^{- 3} \cos (x^{- 4})}{1}

t h e f i r s t l i m i t c o n v e r g e s t o z e r o b y

s q u e e z e t e s t, b u t t h e s e c o n d l i m i t

i s d i v e r g e n t (b e c a u s e x^{- 3} g o e s t o

i n f i n i t y a s x \to 0 a n d \cos (x^{- 4}) d o e s

n o t g o t o z e r o .

\lim_{x \to 0} \frac{2 x \sin (x^{- 4}) - 4 x^{- 3} \cos (x^{- 4})}{1}

d i v e r g e s

Commented by Dwaipayan Shikari last updated on 12/Jul/20

B u t \lim_{x \to 0} \frac{x^{2}}{x} (s i n x^{- 4}) = x \times s i n (x^{- 4}) = 0

a s s i n (x^{- 4}) i s a f i n i t e n u m b e r

Commented by bramlex last updated on 12/Jul/20

this is explanation why

L^{'} Hopital {rule}^{'} s not work to this

equation ⟨ this is Real analysis

∣ \overset{∙}{⌣} ∣ \overset{∙}{⌣} ∣ ⟩

Commented by abdomsup last updated on 12/Jul/20

h o s p i t a l r u l e d o n t w o r k i n t h i s c a s e

b e c a u s e h e i s t i r e d \dots .!

Answered by mathmax by abdo last updated on 12/Jul/20

we have for x \neq 0 \frac{x^{2} \sin (\frac{1}{x^{4}})}{x} = x \sin (\frac{1}{x^{4}}) and ∣ xsin (\frac{1}{x^{4}}) ∣⩽∣ x ∣ \Rightarrow

\lim_{x \to 0} \frac{x^{2} \sin (x^{- 4)}}{x} = 0