Identify the independent variable in an expression [duplicate]

Question

Suppose that you write something like this for your students. This is just the beginning of something I would like to write to help them with the formal definition of a limit, but I am puzzled by the first output.

formalLimit[expr_, a_, L_, ϵ_] := Module[{f},
  f[x_] := expr;
  f[2]
  ]

So, my first question is, when I enter

formalLimit[x^2, 2, 4, .1]

Why do I get the output $x^2$?

Once I get past this problem, my next question is, suppose a student enters an expression $\sqrt{4-t^2}$, not using my choice of $x$ as the independent variable. How can I identify their choice of independent variable in the expr_, then change it to $x$ when using f[x_]:=expr ?

Update: I'd like to thank all of my colleagues for their kind responses. Here is what I was able to do thanks to your help. Students will get a question (yes, it can be answered much more quickly using Mathematica techniques such as Reduce, etc, but I want a visual introduction to the formal definition of a limit) such as "Use a graph to find a number $\delta$ such that

$$|f(x)-L|<\epsilon\qquad\text{whenever}\qquad0<|x-a|<\delta.$$

As an example, find a $\delta$ such that

$$|\sqrt{x}-1|<0.5\qquad\text{whenever}\qquad0<|x-1|<\delta$$

I've written this function:

formalLimit[expr_, var_, a_, L_, ϵ_] := 
  DynamicModule[{f, δ},
    f = (expr /. var -> #) &;
    Manipulate[
      δ = Min[Abs[p[[1]] - a], Abs[q[[1]] - a]];
      Show[
        Plot[f[x], {x, a - zoom, a + zoom}, 
          PlotRange -> {{a - zoom, a + zoom}, {L - 2 ϵ, L + 2 ϵ}},
          Epilog -> 
            {Arrowheads[0.02],
             Arrow[{{a, L - 2 ϵ}, {a, f[a]}, {a - 1.5 δ, L}}],
             {Red, Dashed,
               InfiniteLine[{{p[[1]], L - ϵ}, {p[[1]], L + ϵ}}],
               InfiniteLine[{{q[[1]], L - ϵ}, {q[[1]], L + ϵ}}]}},
          AxesLabel -> {ToString[var], "y"},
          PlotLabel -> "δ = " <> ToString[δ],
          AxesOrigin -> {a - 1.5 δ, L - 2 ϵ},
          Ticks -> {{p[[1]], a, q[[1]]}, {L - ϵ, L, L + ϵ}}],
        Plot[{L - ϵ, L + ϵ}, {x, a - zoom, a + zoom},
          PlotStyle -> Directive[Dashed, GrayLevel[0.8]],
          Filling -> {1 -> {2}}]],
     {{zoom, 3}, 0.0001, 3, Appearance -> "Labeled"},
     {{p, {a - 1, L - ϵ}}, Locator, Appearance -> None},
     {{q, {a + 1, L + ϵ}}, Locator, Appearance -> None}]]

Then the students can enter:

formalLimit[Sqrt[x], x, 1, 1, .5]

And they get this image, where they can use their mouse to drag the dashed vertical lines to help determine $\delta$. There is also a slider for some zooming to adjust the horizontal size of the window.

This function has only been slightly tested, so I wouldn't mind hearing warnings and suggestions.

Answer 1

As Daniel Lichtblau points out in the comments, it's a scoping problem. From the documentation:

The named formal parameters xi in Function[{x1,...}, body] are treated as local, and are renamed xi$ when necessary to avoid confusion with actual arguments supplied to the function.

You can get around this in several ways:

1. Although I don't recommend it, you can explicitly supply x$ as the variable in expr:

   formalLimit[x$^2, 2, 4, .1]
   

   4

   This will not solve your second problem.

2. A better way is to rewrite formalLimit to take an extra argument specifying the variable for which the limit is being taken:

   formalLimitMk2[expr_, var_, a_, L_, \[Epsilon]_] := Module[{f},
     f = (expr /. var -> Slot[]) &;
     f[2]
    ]
   
   formalLimitMk2[x^2, x, 2, 4, .1]
   

   4

3. If your functions will all have only one variable, you can use Variables to extract it:

   formalLimitMk3[expr_, a_, L_, \[Epsilon]_] := Module[{f},
     f = (expr /. First@Variables@expr -> Slot[]) &;
     f[2]
    ]
   
   formalLimitMk3[Sqrt[4 - t^2], x, 2, 4, .1]
   

   0

   Variables is described in the documentation as working for polynomials.

Answer 2

Use an optional argument for the variable with a default of x

Clear[formalLimit]

formalLimit[expr_, a_, L_, ϵ_, sym : _Symbol : x] :=
 Module[{f}, f[y_] := expr /. sym -> y;
  f[2]]


formalLimit[x^2, 2, 4, .1]

(*  4  *)

formalLimit[z^2, 2, 4, .1, z]

(*  4  *)

Answer 3

I would approach of problem of giving students a formal limit evaluator from a different point of view. I would write a multivariate formal limit evaluator and specialize for the single variable case. Since I have a version of Mathematica that is later than V10.1, I would make use of the fairly new ContainsAll function.

formalLimit[expr_, x_Symbol -> val_] := formalLimit[expr, {x -> val}]
formalLimit[expr_, rules : {(_Symbol -> _) ..}] :=
  (expr /. rules) /; ContainsAll[Cases[expr, _, ∞], rules[[All, 1]]]

With these definitions, the following evaluations succeed.

formalLimit[1/x, x -> ∞]

0

formalLimit[Sqrt[x^2 + y^2], {x -> 3, y -> 4}]

5

But these evaluations fail as they should,

formalLimit[Sqrt[x^2 + z^2], {x -> 3, y -> 4}]

formalLimit[Sqrt[x^2 + z^2], {x -> 3, y -> 4}]

formalLimit[Sqrt[x^2 + y^2], {x -> 3, z -> 4}]

formalLimit[Sqrt[x^2 + y^2], {x -> 3, z -> 4}]

With[{x = 42}, formalLimit[Sqrt[x^2 + y^2], {x -> 3, y -> 4}]]

formalLimit[Sqrt[1764 + y^2], {42 -> 3, y -> 4}]

Answer 4

This solution works for any variable identifier consisting of one character

formalLimit[expr_, a_, L_, eps_] :=
  Block[{f, x$, p},
    p = Position[expr, s_Symbol /; StringLength@ToString@s == 1];
    f[x_] := ReplacePart[expr, p -> x$];
    f[2]]

formalLimit[Sqrt[4 - t^2], 2, 4, .1]

0