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I have more complex functions than this, but say I just have:

f[x_]:=(x-a)/(x-2)

The function is not defined for x==2. I would like to find the value of the parameter a, so that I have only a removable discontinuity there, and not an essential (or infinite) one. Clearly, for this simple example, I want a to equal 2. I searched around quite a bit, but came up empty-handed.

My first thought was to require that the two one-sided limits would equal each other, i.e.:

Reduce[Limit[f[x], x -> 2, Direction -> "FromBelow"] ==
    Limit[f[x], x -> 2, Direction -> "FromAbove"], {a}, Reals]

however the result is:

Reduce::infc: The system (-2+a) ∞ == (-2+a) (-∞) contains
    an infinite object ∞.

The problem results from the fact that the Limit[]s happen before Reduce[] gets a chance to try out various values of a. At this point, I could cheat, replacing the ∞ with 1 and -∞ with -1, and then feed it to Reduce[]. But, that would be cheating, and also would only work for this specific case. FindInstance[] was no further help.

NLimit[] (with Chop[] if necessary) from the NumericalCalculus package would not work, as it requires the function to be numeric -- not the case here with the parameter a.

Moving on, I thought of going back to the definition of a limit. The following is shamelessly (shamefully?) snarfed and adjusted from the "Neat Example" at the end of the Reduce reference page:

Reduce[ForAll[ϵ, ϵ > 0,
    Exists[δ, δ > 0,
    ForAll[x, Element[x | a, Reals] && 0 < x - 2 < δ,
    RealAbs[f[x] - f[4 - x]] < ϵ]]], a]

(Trust me, it looks a lot nicer with all of the Greek characters and mathematical symbols in there. :-) )

This nicely did the job, producing: a==2. This even worked for a (only slightly?) more complex:

g[x_] := f[x]^4 - f[x]^2 + 1

However, it did not work for:

h[x_] := Sin[f[x]]

So, my question is: does anyone have an idea how to get this to work for more general functions with parameters, with either of the above forms of Reduce[], or perhaps another approach?

Thank you all for your time and ideas.

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You could do a series expansion, find all the poles, and set them to 0. Here is a function that does this:

removable[expr_, x_->v_, param_] := Module[{poles},
    poles = Cases[
        Series[expr, {x, v, -1}],
        HoldPattern@SeriesData[a__, max_, 1] :> SeriesData[a, 0, 1][[3]],
        {0, -1}
    ];
    Solve[Flatten @ poles == 0, param]
]

For your examples:

removable[(x-a)/(x-2), x->2, a]
removable[((x-a)/(x-2))^4-((x-a)/(x-2))^2+1, x->2, a]
removable[Sin[(x-a)/(x-2)], x->2, a]

{{a -> 2}}

{{a -> 2}}

{{a -> 2}}

A slightly more complicated example:

removable[(x^2-a x-2a^2)/(x-2) + Sin[(x-b^2)/(x-2)], x -> 2, {a, b}]

{{a -> -2, b -> -Sqrt[2]}, {a -> 1, b -> -Sqrt[2]}, {a -> -2, b -> Sqrt[2]}, {a -> 1, b -> Sqrt[2]}}

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  • $\begingroup$ Carl, thank you very much for this answer. I should have pointed out that I am teaching Calculus I, and therefore would like to stick to the relevant Calculus I mathematics of limits and continuity. In other words, I would like to solve this problem with more Mathematica, rather than more mathematics. Any ideas? $\endgroup$ – Aharon Naiman Oct 26 '18 at 7:29
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One way to find the location of the discontinuities is to find the domain of the function. For your example, this gives:

f[x_] := (x - a)/(x - 2);
FunctionDomain[f[x], x]
x < 2 || x > 2

This also works for your other examples:

g[x_] := f[x]^4 - f[x]^2 + 1;
h[x_] := Sin[f[x]];
{FunctionDomain[g[x], x], FunctionDomain[h[x], x]}
{x < 2 || x > 2, x < 2 || x > 2}
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  • 2
    $\begingroup$ I think the OP wanted to extract the value of a, not the location of the discontinuity. They aren't necessarily the same thing, e.g., (x-a-1)/(x-2) $\endgroup$ – Carl Woll Oct 25 '18 at 22:44
  • $\begingroup$ Carl, thank you for pointing this out. Sorry if I was not clear enough. $\endgroup$ – Aharon Naiman Oct 26 '18 at 7:25

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