Seeing double
( x , y ) → ( 0 , 0 ) lim x 2 + y 8 x y 4 = ?
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3 solutions
Discussions for this problem are now closed
I think we can also see this by writing it y 4 x + x y 4 1 If x = y 4 then limit is 0 otherwise 2 1
Cool ! This observation says also l i m i n f = − 2 1 l i m s u p = 2 1
Good observation, but note Andrea Palma's solution. In general, along the path x = k y 4 for any real k the limit would be k 2 + 1 k . Thus the limit can be any real value from − 2 1 to 2 1 inclusive.
I suppose the limit would depend on how fast they are tending to zero relative to each other
Can you please explain how could one tends faster relative to other?
Yes ! More generally it depends more on paths you select. Consider for istance f ( x , y ) = x ∣ x ∣ + 2 y ∣ y ∣ .
Yeah, Paul's Online Math Notes is one of the best sites out there. :)
And now as you have said that. I need to find ways to download the whole site - This is a better example of ambition than what my dictionary has. :)
@Soumo Mukherjee – Download them, digest them, then upload them to the Brilliant Wiki! :)
Sir , why have we considered the path more specifically here , as for any other limit except zero and one , one should assume that two independent variables are approaching the limit with same ' speed ' or linearly as default ! , (until any such constraint has been given over the variables dependency) , hence for any other limit (except zero in such cases ) we get a constant , and with this way only i marked zero . More emphasized with the Shivang's case of obtaining zero for any case when x not equals y^4 and vice versa.
In limits involving one variable we can only approach the limit point from two directions, namely from the left or the right. With limits involving two variables we can approach the limit point from an infinite number of directions, and for the limit to exist it must have the same value no matter which path we approach the limit point along. So if we can find at least two paths along which we find different limits, then we are forced to conclude that the limit does not exist.
For a more complete explanation have a look at this link .
For these two curves x = y 4 and x = 2 y 4 (both passing through ( 0 ; 0 ) ) the function is constantly 2 1 and 5 2 respectively. So the limit doesn't exist.
More generally if you substitute y=mx the value of the limit depends on the value of m. So the limit doesn't exist
This case is nice: for every m ∈ R
x ↦ 0 lim f ( x , m x ) = 0
so ∇ f = ( 0 , 0 ) = so every directional derivative is 0 . Yet the limist doesn't exist because there are other paths to be considered.
When (x,y) ->(0- ,0-) numerator is negative while denominator is positive while when (x,y)->(0+,0+) both numerator and denominator are positive. Hence left hand limit does not coincide with right hand limit ,thus limit does not exist.
Moderator note:
This solution has been marked wrong. You couldn't apply the double sided limits when the limits are combined together as one: ( x , y ) → ( 0 , 0 ) lim .
First look at the limit along a path y = m x for any non-zero real m . The limit is then
( x , y → 0 , 0 ) lim x 2 + m 8 x 8 m 4 x 5 = ( x , y → 0 , 0 ) lim 1 + m 8 x 6 m 4 x 3 = 0 .
Now look at the limit along the path x = y 4 . The limit is then
( x , y → 0 , 0 ) lim 2 y 8 y 8 = 2 1 .
Since the limit is path-dependent, the limit does not in fact exist.
Comment: WolframAlpha gives a result of 0 , which is incorrect. WolframAlpha is a great tool, but I have found it to be completely unreliable when it comes to limits involving two variables. It is only right about half the time.