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]
