To solve the limit [tex]\(\operatorname{Lt}_{x \rightarrow \infty} \frac{(2 x-3)(3 x-4)}{(4 x-5)(5 x-6)}\)[/tex], let's break it down step by step.
1. Write down the given expression:
[tex]\[
\frac{(2x - 3)(3x - 4)}{(4x - 5)(5x - 6)}
\][/tex]
2. Expand the numerator and the denominator:
[tex]\[
\text{Numerator: } (2x - 3)(3x - 4) = 2x \cdot 3x + 2x \cdot (-4) + (-3) \cdot 3x + (-3) \cdot (-4)
= 6x^2 - 8x - 9x + 12 = 6x^2 - 17x + 12
\][/tex]
[tex]\[
\text{Denominator: } (4x - 5)(5x - 6) = 4x \cdot 5x + 4x \cdot (-6) + (-5) \cdot 5x + (-5) \cdot (-6)
= 20x^2 - 24x - 25x + 30 = 20x^2 - 49x + 30
\][/tex]
3. Formulate the limit with the expanded numerator and denominator:
[tex]\[
\operatorname{Lt}_{x \rightarrow \infty} \frac{6x^2 - 17x + 12}{20x^2 - 49x + 30}
\][/tex]
4. Divide every term in the numerator and denominator by [tex]\(x^2\)[/tex], the highest power of [tex]\(x\)[/tex] present:
[tex]\[
\operatorname{Lt}_{x \rightarrow \infty} \frac{6 - \frac{17}{x} + \frac{12}{x^2}}{20 - \frac{49}{x} + \frac{30}{x^2}}
\][/tex]
5. As [tex]\(x\)[/tex] approaches infinity, the terms [tex]\(\frac{17}{x}\)[/tex], [tex]\(\frac{12}{x^2}\)[/tex], [tex]\(\frac{49}{x}\)[/tex], and [tex]\(\frac{30}{x^2}\)[/tex] all approach 0:
[tex]\[
\operatorname{Lt}_{x \rightarrow \infty} \frac{6 - 0 + 0}{20 - 0 + 0} = \frac{6}{20} = \frac{3}{10}
\][/tex]
Thus, the limit is:
[tex]\[
\boxed{\frac{3}{10}}
\][/tex]
