无需洛必达法则的极限替代解法：多项式除法推导疑问

Hey there! Let's work through this limit step by step to get that finite answer you know exists. First, let's restate the original limit to ground our work:
$$\lim_{x \to a} \frac{x}{x-a} \left( \frac{x^3}{(a-1)2} - \frac{a^3}{(x-1)2} \right)$$

Step 1: Simplify the Expression #

First, combine the fractions inside the parentheses over a common denominator, then factor out terms that approach finite values as $x \to a$ (since those won't contribute to the indeterminate $0/0$ form):
$$
\begin{align*}
&= \lim_{x \to a} \frac{x}{x-a} \cdot \frac{x^3(x-1)2 - a^3(a-1)2}{(a-1)^2(x-1)2} \
&= \frac{a}{(a-1)^2 \cdot (a-1)^2} \cdot \lim_{x \to a} \frac{x^3(x-1)2 - a^3(a-1)2}{x - a}
\end{align*}
$$
Here, we've substituted $x \to a$ into the terms $x$ and $(x-1)^2$ directly because they don't create an indeterminate form—they just approach $a$ and $(a-1)^2$ respectively.

Step 2: Recognize the Derivative Definition #

The remaining limit is exactly the definition of the derivative of $f(x) = x^3(x-1)2$ at $x=a$:
$$\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$$
This is way simpler than polynomial long division! Let's compute $f'(x)$:

First, expand $f(x)$ for easier differentiation:
$f(x) = x^3(x2 - 2x + 1) = x^5 - 2x^4 + x^3$ (matches the numerator you expanded earlier, by the way!)

Take the derivative using basic power rule:
$f'(x) = 5x^4 - 8x^3 + 3x^2$

Evaluate at $x=a$:
$f'(a) = 5a^4 - 8a^3 + 3a^2 = a^2(5a2 - 8a + 3) = a^2(5a - 3)(a - 1)$

Step 3: Combine and Simplify #

Plug $f'(a)$ back into our expression and simplify:
$$
= \frac{a}{(a-1)^4} \cdot a^2(5a - 3)(a - 1) = \frac{a^3(5a - 3)}{(a-1)^3}
$$

Quick Note on Your Polynomial Division Issue #

When you tried dividing the expanded numerator by $(x-a)$, you should have gotten a remainder of 0 (since $x=a$ makes the numerator zero: $a^5 -2a^4 +a^3 -a^5 +2a^4 -a^3=0$). The "infinite remainder" you saw was likely an arithmetic error in the division steps. Using the derivative definition (or L'Hospital's Rule on the $0/0$ form) is a far more efficient path here.

内容的提问来源于stack exchange，提问作者Ariel Arévalo