Alternate solution to a limit without using L'Hopital's rule

Question

$$\lim \limits_{x \to a} \frac{x}{x-a} \left(\frac{x^3}{(a-1)^2}-\frac{a^3}{(x-1)^2}\right)$$

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.

Answer 1

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?

Answer 2

$$\lim \limits_{x \to a} \frac{x}{x-a} \left(\frac{x^3}{(a-1)^2}-\frac{a^3}{(x-1)^2}\right)$$

$$=\lim_{x\to a}x\left(\dfrac1{(a-1)^2}\lim_{x\to a}\dfrac{x^3-a^3}{x-a}+a^3\lim_{x\to a}\dfrac{\dfrac1{(a-1)^2}-\dfrac1{(x-1)^2}}{x-a}\right)$$

Now $\lim_{x\to a}\dfrac{\dfrac1{(a-1)^2}-\dfrac1{(x-1)^2}}{x-a}$

$=\dfrac1{(a-1)^2}\lim_{x\to a}\dfrac1{(x-1)^2}\lim_{x\to a}\dfrac{(x-1)^2-(a-1)^2}{x-a}=?$