b. $$ \quad \lim _{x \rightarrow \infty} \frac{3-2 x}{3 x^{3}-1} $$

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Calculate the limit $$ \lim_{x\rightarrow \infty} \frac{3-2x}{3x^3-1} $$.
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
  $$\lim _{x\rightarrow +\infty}\left(\frac{3-2x}{3x^{3}-1}\right)$$
- step1: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(\frac{x\left(\frac{3}{x}-2\right)}{x^{3}\left(3-\frac{1}{x^{3}}\right)}\right)$$
- step2: Reduce the fraction:
  $$\lim _{x\rightarrow +\infty}\left(\frac{\frac{3}{x}-2}{x^{2}\left(3-\frac{1}{x^{3}}\right)}\right)$$
- step3: Rewrite the expression:
  $$\frac{\lim _{x\rightarrow +\infty}\left(\frac{3}{x}-2\right)}{\lim _{x\rightarrow +\infty}\left(x^{2}\left(3-\frac{1}{x^{3}}\right)\right)}$$
- step4: Calculate:
  $$\frac{-2}{\lim _{x\rightarrow +\infty}\left(x^{2}\left(3-\frac{1}{x^{3}}\right)\right)}$$
- step5: Calculate:
  $$\frac{-2}{+\infty}$$
- step6: Calculate:
  $$0$$
The limit of the expression $$ \frac{3-2x}{3x^3-1} $$ as $$ x $$ approaches infinity is 0.
The limit is 0.

b.

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