find lim x→0 sin 3x / (5x² - 4x)

Explanation:

Step1: Use the limit - formula $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$

We can rewrite the given limit $\lim_{x\rightarrow0}\frac{\sin3x}{5x^{2}-4x}$ as $\lim_{x\rightarrow0}\frac{\sin3x}{x(5x - 4)}$. Let $u = 3x$, then as $x\rightarrow0$, $u\rightarrow0$. And $\sin3x=\sin u$, and we can rewrite the limit as $\lim_{x\rightarrow0}\frac{\sin3x}{3x}\cdot\frac{3}{5x - 4}$.

Step2: Apply the limit rules

We know that $\lim_{x\rightarrow0}\frac{\sin3x}{3x}=1$. Then, by the product - rule of limits $\lim_{x\rightarrow a}(f(x)\cdot g(x))=\lim_{x\rightarrow a}f(x)\cdot\lim_{x\rightarrow a}g(x)$ (where $f(x)=\frac{\sin3x}{3x}$ and $g(x)=\frac{3}{5x - 4}$), we have: [ \begin{align*} \lim_{x\rightarrow0}\frac{\sin3x}{3x}\cdot\frac{3}{5x - 4}&=\lim_{x\rightarrow0}\frac{\sin3x}{3x}\cdot\lim_{x\rightarrow0}\frac{3}{5x - 4}\ &=1\cdot\frac{3}{5\times0 - 4} \end{align*} ]

Step3: Calculate the final result

[1\cdot\frac{3}{0 - 4}=-\frac{3}{4}]

Answer:

$-\frac{3}{4}$