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

Question

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. $$ \lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2} = \lim_{x \to 3} \frac{x^{14} + 3x^{13} + ... + 3^{14} - 15 \cdot 3^{14}}{x-3} $$

Answer 1

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 $0$.

$\require{cancel}\begin{align}f(x)&=\dfrac{(3+u)^{15}-3^{15}-15\cdot3^{14}u}{u^2}\\\\&=\dfrac{\bigg(\cancel{3^{15}}+\cancel{15\cdot3^{14}u}+\binom{15}{2}3^{13}u^2+o(u^2)\bigg)-\cancel{3^{15}}-\cancel{15\cdot3^{14}u}}{u^2}\\\\&=3^{13}\times\frac{14\times15}{2}+o(1)\to 35\times 3^{14}\end{align}$

Answer 2

Now, $$\frac{x^{14}+3x^{13}+...+3^{13}x-14\cdot3^{14}}{x-3}=\frac{x^{14}-3^{14}}{x-3}+3\cdot\frac{x^{13}-3^{13}}{x-3}+...+3^{13}\rightarrow$$ $$=14\cdot3^{13}+3\cdot13\cdot3^{12}+...+3^{13}=(14+13+...+1)3^{13}=105\cdot3^{13}=35\cdot3^{14}.$$ I used the following: $$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}\right)$$ for natural $n\geq2$.

We see that in the big brackets we have $n$ terms $a^n$ for $a=b$.