How to calculate it without l'Hopital?

Question

How to calculate this limit without using l'Hopital rule? $$\lim_{x\to 1 }\frac{3^{5x}-3^{2x^2+3}}{\sin(\pi x)}$$ I know how to make it using L'Hopital and that the result is $-\frac{243\log(3)}{\pi}$, but I'm getting nowhere when I try without it. Any advice? I also tried to change ${\sin(\pi x)}$ to ${\tan x}*{\cos x}$, but it didn't help.

Answer 1

Hint:

Set $x-1=h$

$$\lim_{h\to0}\dfrac{3^{5+5h}-3^{2(h+1)^2+3}}{\sin\pi(1+h)}$$

$$=-\dfrac{5\cdot3^5\lim_{h\to0}\dfrac{3^{5h}-1}{5h}-3^5\lim_{h\to0}\dfrac{3^{2h^2+4h}-1}{2h^2+4h}\cdot\lim_{h\to0}\dfrac{2h^2+4h}h}{\pi\cdot\lim_{h\to0}\dfrac{\sin\pi h}{\pi h}}$$

Now use for $a>0,$ $$\lim_{h\to0}\dfrac{a^h-1}h=\ln a$$