algebraic manipulation - Limiting form of a polynomial expression

When simplifying an expression by hand, a trick that is often used is to remove terms that are lower powers of the independent variable, for instance, as $x \rightarrow \infty$,

$x^2 + x$

becomes simply

$x^2$

I am trying to replicate this functionality in Mathematica. Limit seems like an obvious choice, but

Limit[x^2 + x, x -> ∞]

simply returns $∞$ (not terribly helpful). Another option is Series, but it seems that one has to pass the highest power in the expression to get the desired result, with

Normal@Series[x^2 + x, {x, ∞, -2}]

returning the expected $x^2$. However, this does not work in general for more complicated expressions, when the highest power may not be so obvious. Consider the following example:

$\frac{x^2+x}{x+1}$

Mathematica code

(x^2+x)/(x+1)

We expect that the limiting form of this expression as $x \rightarrow \infty$ should be $x$ (the top becomes $x^2$, the bottom becomes $x$, and this then simplifies to $x$). However, if you just use the highest power available, Series returns 0 (you have to actually use Normal@Series[x^2 + x, {x, ∞, -1}]).

My question is, can anyone produce a generalized piece of code that can produce a limiting expression? It would be good if the procedure handled the other extreme as well, when the variable in question approaches 0 (in which case $x^2 + x$ would simplify to $x$).

Edit: It seems that the main issue for the solutions presented so far is that the form of the expression is not consistent. These operations are relatively easy to perform by hand, but sometimes tedious, which is why it would be nice to have Mathematica do it for me! Here's a pathological example that I can reduce by hand and for which the accepted solution should be able to return the correct result:

$\frac{\sqrt{x^3 + x}}{x^5 + x + 1} \frac{x^2}{x^{10} + x} + \frac{x^5 + x}{x}$

Mathematica code

Sqrt[x^3 + x]/(x^5 + x + 1) x^2/(x^10 + x) + (x^5 + x)/x

(The right answer is $x^4$)

Answer

The solution to this requires just a minor modification of my answer here:

ClearAll[limitingTerm1]
SetAttributes[limitingTerm1, HoldAll]
limitingTerm1[expr_, var_Symbol, func_] := var^Exponent[#, var, func] &@Simplify[expr]

This gives the right answer for both the examples provided by the OP:

limitingTerm1[(x^2 + x)/(x + 1), x, Max]
(* x *)


limitingTerm1[Sqrt[x^3 + x]/(x^5 + x + 1) x^2/(x^10 + x) + (x^5 + x)/x, x, Max]
(* x^4 *)

---

One can now generalize this to polynomials with arbitrary coefficients. It works as a two-pass approach, where the second pass is applied to the result of the first if its denominator still contains an x term. This seems to work very well in examples I've tested, but it might be possible to find edge cases where it doesn't work. Nevertheless, some modifications and extensions along these lines should help solve it.

ClearAll[limitingTerm2]
SetAttributes[limitingTerm2, HoldAll]
limitingTerm2[expr_, var_Symbol, func_] := Module[{term},
    term := With[{pow = Exponent[#, var, func]}, Coefficient[#, var, pow] var^pow] &;

    If[FreeQ[Denominator@#, x], #, term@Apart@#] &@term@Simplify[expr]
]

limitingTerm2[(a x^2 + bx)/(d + c x), x, Max]
(* (a x)/c *)

It also gives the correct answer for the other examples above.