Limit, two variables

  • #1
209
1
In my calculus book it says that the limit of y(x^3) as (x,y)->(0,1) equals 0. It also says that a limit does not exist if you obtain different values when approaching (0,1) from different paths.
It is easy to see the limit is zero by using the product rule for limits. However, if I set x=y, we get the limit of x^4 as (x,y)->(0,1) or the limit of y^4 as (x,y)->(0,1) which are clearly not equal. Hence the limit does not exist (?).

Where is my reasoning false?
 

Answers and Replies

  • #2
sylas
Science Advisor
1,647
7
In my calculus book it says that the limit of y(x^3) as (x,y)->(0,1) equals 0. It also says that a limit does not exist if you obtain different values when approaching (0,1) from different paths.
It is easy to see the limit is zero by using the product rule for limits. However, if I set x=y, we get the limit of x^4 as (x,y)->(0,1) or the limit of y^4 as (x,y)->(0,1) which are clearly not equal. Hence the limit does not exist (?).

Where is my reasoning false?

By thinking that you can get to (0,1) along a path where x=y?

Cheers -- sylas
 
  • #3
209
1
Haha, how stupid is that! :)

So I could substitute x=y-1, then the limit becomes (y-1)y^4 as (x,y)->(0,1) which DOES equal zero and everything works out. Thanks!
 
  • #4
sylas
Science Advisor
1,647
7
Haha, how stupid is that! :)

So I could substitute x=y-1, then the limit becomes (y-1)y^4 as (x,y)->(0,1) which DOES equal zero and everything works out. Thanks!

You're welcome. Always glad to contribute another data point on the war between engineers and scientists. :biggrin:
 

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