Answer :

Let's evaluate the limit [tex]\(\lim_{x \to -\infty} \frac{4 - 5x + 6x^2}{(3x - 1)^2}\)[/tex].

To do this, we go through the following steps:

1. Identify the dominant terms in the numerator and the denominator.

For large values of [tex]\(x\)[/tex] (specifically as [tex]\(x \to -\infty\)[/tex]), the contributions of the highest power of [tex]\(x\)[/tex] will dominate. Here, the highest power of [tex]\(x\)[/tex] in the numerator is [tex]\(6x^2\)[/tex] and in the denominator is [tex]\((3x)^2\)[/tex].

2. Rewrite the expression using these dominant terms.

The numerator [tex]\(4 - 5x + 6x^2 \approx 6x^2\)[/tex] as [tex]\(x\)[/tex] becomes very large (negatively).

The denominator [tex]\((3x - 1)^2 \approx (3x)^2\)[/tex] as [tex]\(x\)[/tex] becomes very large (negatively).

3. Simplify the expression.

So, for large [tex]\(x\)[/tex]:

[tex]\[ \frac{4 - 5x + 6x^2}{(3x - 1)^2} \approx \frac{6x^2}{(3x)^2} \][/tex]

4. Simplify the fraction.

[tex]\[ \frac{6x^2}{(3x)^2} = \frac{6x^2}{9x^2} = \frac{6}{9} = \frac{2}{3} \][/tex]

Thus, as [tex]\(x \to -\infty\)[/tex], the fraction [tex]\(\frac{6x^2}{(3x)^2}\)[/tex] simplifies to [tex]\(\frac{2}{3}\)[/tex].

So, we conclude that:

[tex]\[ \lim_{x \to -\infty} \frac{4 - 5x + 6x^2}{(3x - 1)^2} = \frac{2}{3} \][/tex]

Therefore, the final result is:

[tex]\[ \boxed{\frac{2}{3}} \][/tex]