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

lim_(x→0) ((x^2  sin (x^(−4) ))/x) ?

Answered by bemath last updated on 12/Jul/20

lim_(x→0) ((2x sin (x^(−4) )−4x^(−3)  cos (x^(−4) ))/1)  the first limit converges to zero by  squeeze test , but the second limit  is divergent (because x^(−3)  goes to  infinity as x→0 and cos (x^(−4) ) does  not go to zero.  lim_(x→0) ((2x sin (x^(−4) )−4x^(−3)  cos (x^(−4) ))/1)  diverges

Commented byDwaipayan Shikari last updated on 12/Jul/20

But lim_(x→0)  (x^2 /x)(sin x^(−4) )=x×sin(x^(−4) )=0  as sin(x^(−4) ) is a finite number

Commented bybramlex last updated on 12/Jul/20

this is explanation why  L′Hopital rule′s not work to this  equation ⟨this is Real analysis  ∣⌣^• ∣⌣^• ∣ ⟩

Commented byabdomsup last updated on 12/Jul/20

hospital rule dont work in this case  because he is tired....!

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

we have  for x≠0   ((x^2  sin((1/x^4 )))/x) =x sin((1/x^4 ))  and ∣xsin((1/x^4 ))∣≤∣x∣ ⇒  lim_(x→0)     ((x^2  sin(x^(−4)) )/x) =0