Let f(x,y)=y/(1+xy)−(1−ysin(pi xy))/(arctan(x)) Compute the limit: g(x)=lim_(y->oo) f(x,y) Wouldn't the value of the limit also depend on x?

Widersinnby7

Widersinnby7

Answered question

2022-11-06

Let f ( x , y ) = y 1 + x y 1 y sin ( π x y ) arctan ( x )
Compute the limit:
g ( x ) = lim y f ( x , y )
Wouldn't the value of the limit also depend on x?

Answer & Explanation

kuthiwenihca

kuthiwenihca

Beginner2022-11-07Added 23 answers

Since you are having problems with the second term, note that you can write it as
1 sin π x p p t a n 1 x
(here 1 p = y and as y goes to , p goes to 0) leaving the t a n 1 x alone, our job is to find
lim p 0 sin π x p p = lim p 0 sin π x p π x p π x= π x
Sophie Marks

Sophie Marks

Beginner2022-11-08Added 2 answers

Let f ( x , y ) = y 1 + x y 1 y s i n ( π x y ) arctan ( x ) , let's modify our function
y 1 + x y 1 y s i n ( π x y ) arctan ( x ) = 1 x 1 1 x y + 1 1 y s i n ( π x y ) arctan ( x ) ,
now let z = 1 y , hence
lim y f ( x , y ) = lim z 0 f ( x , 1 z ) = lim z 0 1 x 1 z x + 1 1 1 z s i n ( z π x ) arctan ( x ) ,
clearly 1 x 1 z x + 1 1 x , let's have a closer look at 1 1 z s i n ( z π x ) arctan ( x ) .
Using Taylor series we can write
1 z s i n ( z π x ) = 1 z ( z π x + O ( z 3 ) ) = π x + O ( z 2 ) ,
Can you go from here?

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