Evaluate the limit $$ \lim _{x \rightarrow \infty} \frac{(2-x)(7+2 x)}{(3-3 x)(9+3 x)} $$

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Calculate the limit $$ \lim _{x\rightarrow \infty} \frac{(2-x)(7+2x)}{(3-3x)(9+3x)} $$.
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
  $$\lim _{x\rightarrow +\infty}\left(\frac{\left(2-x\right)\left(7+2x\right)}{\left(3-3x\right)\left(9+3x\right)}\right)$$
- step1: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(\frac{x^{2}\left(\frac{14}{x^{2}}-\frac{3}{x}-2\right)}{x^{2}\left(\frac{27}{x^{2}}-\frac{18}{x}-9\right)}\right)$$
- step2: Reduce the fraction:
  $$\lim _{x\rightarrow +\infty}\left(\frac{\frac{14}{x^{2}}-\frac{3}{x}-2}{\frac{27}{x^{2}}-\frac{18}{x}-9}\right)$$
- step3: Rewrite the expression:
  $$\frac{\lim _{x\rightarrow +\infty}\left(\frac{14}{x^{2}}-\frac{3}{x}-2\right)}{\lim _{x\rightarrow +\infty}\left(\frac{27}{x^{2}}-\frac{18}{x}-9\right)}$$
- step4: Calculate:
  $$\frac{-2}{\lim _{x\rightarrow +\infty}\left(\frac{27}{x^{2}}-\frac{18}{x}-9\right)}$$
- step5: Calculate:
  $$\frac{-2}{-9}$$
- step6: Calculate:
  $$\frac{2}{9}$$
The limit of the expression $$ \frac{(2-x)(7+2x)}{(3-3x)(9+3x)} $$ as $$ x $$ approaches infinity is $$ \frac{2}{9} $$ or approximately 0.222.
The limit is $$ \frac{2}{9} $$ or approximately 0.222.

Evaluate the limit

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