Answer
$$\lim\limits_{x\to0}\frac{\sin 3x}{5x^3-4x}=-\frac{3}{4}$$
Work Step by Step
$$A=\lim\limits_{x\to0}\frac{\sin 3x}{5x^3-4x}$$
Again, try to change the formula into a new form having $\frac{\sin x}{x}$ or $\frac{\cos x-1}{x}$. $$A=\lim\limits_{x\to0}\Bigg(\frac{\sin 3x}{3x}\times\frac{3x}{5x^3-4x}\Bigg)$$ $$A=\lim\limits_{x\to0}\frac{\sin 3x}{3x}\times\lim\limits_{x\to0}\frac{3x}{5x^3-4x}$$ $$A=B\times C$$
1) Consider B: $$B=\lim\limits_{x\to0}\frac{\sin 3x}{3x}$$
Let $3x=\theta$. Then as $x\to0$, $\theta\to0$ as well. So, $$B=\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}=1$$
2) Consider C: $$C=\lim\limits_{x\to0}\frac{3x}{5x^3-4x}$$ $$C=\lim\limits_{x\to0}\frac{3x}{x(5x^2-4)}$$ $$C=\lim\limits_{x\to0}\frac{3}{5x^2-4}$$ $$C=\frac{3}{5\times0^2-4}$$ $$C=-\frac{3}{4}$$
Therefore $$A=1\times(-\frac{3}{4})=-\frac{3}{4}$$
