<s>[INST] Alternate solution to a limit without using L'Hopital's rule $$\lim \limits_{x \to a} \frac{x}{x-a} \left(\frac{x^3}{(a-1)^2}-\frac{a^3}{(x-1)^2}\right)$$
I've gotten to this 
$$a \cdot \lim \limits_{x \to a} \frac{x^5-2x^4+x^3-a^5+2a^4-a^3}{(x-a)(a-1)^2(x-1)^2}$$
since as far as I'm concerned it's just a matter of doing a long polynomial division between the numerator and $(x-a)$, but I'm getting a remainder that equals infinity despite knowing the answer is finite.
Upon looking up the division itself it seems I had made a mistake, but I'm still curious as to how else this could be solved other than essentially brute forcing it.
 [/INST] One way is to solve with Differentiation From First Principles perspective:
$\displaystyle \lim \limits_{x \to a} \frac{x}{x-a} \left(\frac{x^3}{(a-1)^2}-\frac{a^3}{(x-1)^2}\right)=\lim \limits_{x \to a} \frac{x}{x-a} \left(\dfrac{x^3(x-1)^2-a^3 (a-1)^2}{(a-1)^2 (x-1)^2}\right)=\dfrac{a}{(a-1)^4} \left( \lim_{x \to a} \dfrac{x^3(x-1)^2-a^3 (a-1)^2}{x-a}  \right)=\dfrac{a}{(a-1)^4}(x^3(x-1)^2)'|_{x=a}$
Can you take it from here?
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