To evaluate the limit [tex]\(\lim_{x \rightarrow \infty} \frac{(3x-1)^8 \cdot (4x-1)^2}{(3x+1)^{10}}\)[/tex], we can use a step-by-step approach:
1. Factor out the dominant terms: As [tex]\(x\)[/tex] approaches infinity, the terms [tex]\(3x\)[/tex] and [tex]\(4x\)[/tex] will dominate over the constants [tex]\(1\)[/tex] or [tex]\(-1\)[/tex]. Therefore, we can factor these terms out from the numerator and the denominator:
[tex]\[
\frac{(3x-1)^8 \cdot (4x-1)^2}{(3x+1)^{10}} \approx \frac{(3x)^8 \cdot (4x)^2}{(3x)^{10}}.
\][/tex]
2. Simplify the expression: Simplify the exponents:
[tex]\[
\frac{(3x)^8 \cdot (4x)^2}{(3x)^{10}} = \frac{(3^8 \cdot x^8) \cdot (4^2 \cdot x^2)}{(3^{10} \cdot x^{10})}.
\][/tex]
3. Combine the terms in the numerator and denominator:
[tex]\[
= \frac{3^8 \cdot 4^2 \cdot x^{8+2}}{3^{10} \cdot x^{10}} = \frac{3^8 \cdot 16 \cdot x^{10}}{3^{10} \cdot x^{10}}.
\][/tex]
4. Simplify the exponents and coefficients:
[tex]\[
= \frac{3^8 \cdot 16}{3^{10}} = \frac{16}{3^{10-8}} = \frac{16}{3^2} = \frac{16}{9}.
\][/tex]
5. Conclusion:
Hence, the limit is:
[tex]\[
\lim_{x \rightarrow \infty} \frac{(3 x - 1)^8 \cdot (4 x - 1)^2}{(3 x + 1)^{10}} = \frac{16}{9}.
\][/tex]
So, the result of the limit is [tex]\(\boxed{\frac{16}{9}}\)[/tex].
