Revisiting some limits
x y → 1 lim ( ln y ln x + ln x ln y ) = ?
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2 solutions
Discussions for this problem are now closed
Moderator note:
Great!
As an explicit example of the last line, take any point of the form ( u , v ) = ( u , k u ) for a constant k , with u approaching 0. Then, we obtain the value of k + k 1 .
In particular, this shows us that those who obtained a value of - 2, took a path that looked a lot like k = − 1 IE x ≈ − y . It also explains how I constructed my counter example, by letting k be variable and tending off to ∞ .
as XY tends to 1 means (ln X+ln Y)tends to zero. the given function can be written like , {suare of (ln X+ln Y) - 2ln X. ln Y} / ln X. lnY as limit value, the functional value tends to -2.
let L=lnx/lny L+1/L-1=lnxy/lnx-lny which tends to 0 (L!=1) L=-1 similarly the next limit also comes to L=-1 hence limit =-2
It could also be possible if we fix one coordinate, say y=4, and let x approaches 1/4, then the limit will be a constant say k_1, if we change the value of y, say y=5, and follow the same line of thought we will get another limit. Hence, the limit does not exist.
However, is the expression well-defined? I kept wondering about it, what could be the topology of convergence? The incredible answer of changing y=exp{u} gives you a neighbourhood around (0,0), where we can write down the definition of limit, (for every \epsilon>0 there exists....),
Can we do the same thing in the problem as it is written?
If k is "tending off to ∞," how do we know (u, ku) still approaches (0, 0)? We may have "u approaching 0," but it seems that ku would be an indeterminate form and thus no longer valid for being an equivalent limit with the substitutions.
Add 1 to the two fractions to get, ln y ln x + ln y + ln x ln x + ln y − 2 = ln y ln x y + ln x ln x y − 2 = 0 + 0 − 2 = − 2 . And done!
Moderator note:
Always be careful with 2 variable limits. You cannot merely choose a particular path and evaluate the limit along that path. Instead, you have to consider all possible paths, and be really careful. Saying x y → 1 does not mean that we can substitute in x y = 1 whenever we see it.
As an explicit counter-example to this problem, take x = 1 + k 1 , y = 1 − k 2 , with k → 0 . Then, x y = 1 − k → 1 .
However, lim x y → 1 ln y ln x y = lim k → 0 ln ( 1 − k 2 ) ln ( 1 − k ) = lim k → 0 − k 2 − k = ∞ .
We can check that
x y → 1 lim ln x ln x y = k → 0 lim ln ( 1 + k 1 ) ln ( 1 − k ) = k → 0 lim − ln ( 1 + k ) ln ( 1 − k ) = k → 0 lim − k − k = 1 .
Hence, along this path, the limit of the expression is ∞ + 1 = ∞ . Thus, the limit does not exist.
Wrong. You cannot substitute in the value of x y directly.
Sir,can you please add few questions on d o u b l e l i m i t for better understanding of these type of questions.
Thanks!!
I just want to confirm my understanding. In 2 variable limits,limit should exist for all possible solutions of X and Y,only then we can say that the limit exists.
Right, it needs to exist for all possible paths that tend towards that value.
In 1 variable limit, there are only 2 "essentially distinct" paths, namely the Left path, and the Right path. That is why we say "The limit exists if the left hand limit is equal to the right hand limit".
In 2 variables, you need to care about all kinds of crazy paths. We need to be careful with paths in which one variable causes the function to tends off to 0, or inifinity.
@Calvin Lin – thanks for clearing it..
let x = e u and y = e v , we now have to find u + v → 0 lim v u + u v or ( u , v ) → ( u 0 , v 0 ) lim v u + u v with ( u 0 , v 0 ) is a point on the line ( d ) : x + y = 0
We can see that when ( u , v ) approachs ( 0 , 0 ) , the expression can have many values, thus the limit doesn't exist