Limit of x(a,b) as a,b->k: Isolated Variable Method?

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In summary, the conversation discusses the problem of proving that a function x(a,b) is bounded by some value l, with the extremes of the function occurring when a = b = k is approached. The conversation considers whether setting a equal to k and taking the limit of the new function x(b) as b approaches k would be a valid proof, but it is concluded that this is not the case in the general situation. The idea of taking limits over all paths is mentioned. The function x(a,b) is defined and it is stated that the goal is to show that x is in the range of (-k,k). It is suggested that the limit of the function does not exist, and this is confirmed by considering two different paths where the
  • #1
1MileCrash
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I want to show that a function x(a,b) is bounded by some value l, knowing that the extremes of the function occur when a = b = k is approached. The function blows up here.

Being a function of two variables, is it valid proof to set a EQUAL TO k, and then take the limit of this new x(b) as b -> k, getting l as the result?
 
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  • #2
Not knowing the specific situation here my hunch is no. In the general case you definitely need to take limits over all paths (a,b)->(k,k) instead of just checking a specific one.
 
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  • #3
I was under the impression that all of the limits would be the same if the limit exists?
 
  • #4
Ah! If you know a priori that the limit exists, then yes fixing a specific path works.
 
  • #5
Well, I guess I don't, really. I guess I am just making the assumption because the function looked very simple. I have forgotten all about the fact that this is a "path" limit.

Basically, this is the function (k is constant):

x(a,b) = (a-b)/((ba/k^2) - 1)

Where a,b are in (-k, k).
I needed to show that x is in (-k,k) also. Doing algebraic manipulations with inequalities didn't lead me to the result, so I remarked that the functions extremes are when ba = k^2 and so a = k and b = k or a = -k and b = -k, then I proceeded to do as I explained in the OP to show what those values were. I never really considered the possibility that the function's limit did not exist because it seemed so simple (and I think the "almost" symmetry of the function (switching a and b only switches sign) also played a part in thinking I could fix a path.)

I think I need to review some things, though.
 
  • #6
It is easy to see the limit does not exist in that case. For simplicity I will assume k=1 and consider the paths where 1) a=1 and b→1 and 2) a→1 and b=1. In the first case we have x(a,1) = (a-1)/(a-1) = 1 so the limit along this path returns the value 1. In the second case we have x(1,b) = (1-b)/(b-1) = -1 so the limit along this path returns the value -1. If the limit existed, then these two would agree, so evidently the limit does not exist.
 
  • #7
Does there exist some path where the limit is not k or -k?
 
  • #8
1MileCrash said:
Does there exist some path where the limit is not k or -k?

In this case there is such a path. Go along the line a=b as (a,b)→(k,k).
 
  • #9
Alright, it looks like I need to find some other method to show that the range is in (-k, k) then.

Thank you for all the help, do you have any suggestions?

Edit: Please keep any suggestion brief, I would like the joy of solving it.
 
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  • #10
Well I think this function is actually unbounded around (k,k). So hopefully that helps put you on the right track.
 
  • #11
jgens said:
Well I think this function is actually unbounded around (k,k). So hopefully that helps put you on the right track.

Oops! Totally missed that you have a restricted domain -k < a,b < k. In this case your function is indeed bounded and you can prove this essentially by arguing cases. Sorry about that.
 
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What is the "Limit of x(a,b) as a,b->k: Isolated Variable Method"?

The "Limit of x(a,b) as a,b->k: Isolated Variable Method" is a mathematical concept used to determine the behavior of a function as two variables, a and b, approach a common value, k. It is a method of finding the limit by isolating one variable at a time and evaluating the function as that variable approaches the common value.

How is the "Limit of x(a,b) as a,b->k: Isolated Variable Method" different from other methods of finding limits?

The "Limit of x(a,b) as a,b->k: Isolated Variable Method" differs from other methods, such as direct substitution or algebraic manipulation, because it allows for the evaluation of the limit as two variables approach a common value. This can be useful in cases where other methods may not work or may be more complex.

When is the "Limit of x(a,b) as a,b->k: Isolated Variable Method" most commonly used?

The "Limit of x(a,b) as a,b->k: Isolated Variable Method" is commonly used in multivariable calculus and in real-world applications where a function is dependent on multiple variables. It is also useful in cases where the limit cannot be evaluated using other methods.

What are the steps to using the "Limit of x(a,b) as a,b->k: Isolated Variable Method"?

The steps to using the "Limit of x(a,b) as a,b->k: Isolated Variable Method" are as follows:1. Identify the common value, k, that the variables a and b are approaching.2. Isolate one variable at a time by setting the other variable equal to k in the function.3. Evaluate the function as the isolated variable approaches k.4. Repeat for the other variable.5. If both limits exist and are equal, then the limit of the function as x(a,b) approaches (a,b)->k exists and is equal to the common value.

What are some common mistakes when using the "Limit of x(a,b) as a,b->k: Isolated Variable Method"?

Some common mistakes when using the "Limit of x(a,b) as a,b->k: Isolated Variable Method" include:- Forgetting to evaluate both limits separately.- Not setting the other variable equal to k when isolating a variable.- Assuming that the limit exists without checking for continuity.- Not simplifying the function before evaluating the limits.- Not using the correct notation for the limits, such as using parentheses instead of brackets.

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