Find $\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}$ Please help me find:

$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}
$$

I cannot use L'Hospital's rule.
I tried to eliminate $x-3$, but I have no idea what to do next.
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2} = 
\lim_{x \to 3} \frac{x^{14} + 3x^{13} + ... + 3^{14} - 15 \cdot 3^{14}}{x-3}
$$

Let $x=3+u$ with $u\to 0$ then use binomial formula to expand $(3+u)^{15}$.
You can ignore terms in $u^k$ with $k\ge 3$ because when divided by $u^2$ they converge to $0$.
$\require{cancel}\begin{align}f(x)&=\dfrac{(3+u)^{15}-3^{15}-15\cdot3^{14}u}{u^2}\\\\&=\dfrac{\bigg(\cancel{3^{15}}+\cancel{15\cdot3^{14}u}+\binom{15}{2}3^{13}u^2+o(u^2)\bigg)-\cancel{3^{15}}-\cancel{15\cdot3^{14}u}}{u^2}\\\\&=3^{13}\times\frac{14\times15}{2}+o(1)\to 35\times 3^{14}\end{align}$
