Answer :

Alright, let's explore the limit:

[tex]\[ \lim_{x \to \infty} \frac{(3-1)^8 - (4x - y)^2}{(3x + 1)^{10}} \][/tex]

Let's approach this step-by-step:

1. Simplify the constant terms in the numerator:
[tex]\[ (3 - 1)^8 = 2^8 = 256 \][/tex]
Therefore, the numerator becomes:
[tex]\[ 256 - (4x - y)^2 \][/tex]

2. Expand and simplify the expression inside the numerator:
For very large values of [tex]\( x \)[/tex], the term [tex]\( y \)[/tex] becomes negligible in [tex]\( (4x - y)^2 \)[/tex]. Thus, [tex]\( (4x - y)^2 \)[/tex] approximates to [tex]\( (4x)^2 \)[/tex]:
\]
[tex]\[ (4x)^2 = 16x^2 \][/tex]
Hence, the numerator becomes:
[tex]\[ 256 - 16x^2 \][/tex]

3. Simplify the denominator:
[tex]\[ (3x + 1)^{10} \][/tex]
For very large values of [tex]\( x \)[/tex], the term [tex]\( 1 \)[/tex] becomes negligible, making:
\]
[tex]\( (3x + 1)^{10} \approx (3x)^{10} \] Simplifying further, we get: \] (3x)^{10} = 3^{10} \cdot x^{10} = 59049 \cdot x^{10} \] 4. Rewrite the limit expression with the simplified numerator and denominator: \] \( \lim_{x \to \infty} \frac{256 - 16x^2}{59049 \cdot x^{10}} \] 5. Simplify the expression by dividing both the numerator and the denominator by \( x^2 \)[/tex]:
\]
[tex]\( \frac{256 - 16x^2}{59049 \cdot x^{10}} = \frac{256/x^2 - 16x^2/x^2}{59049 \cdot x^{10}/x^2} = \frac{256/x^2 - 16}{59049 \cdot x^8} \] 6. Take the limit as \( x \)[/tex] approaches infinity:
\]
[tex]\( \lim_{x \to \infty} \frac{256/x^2 - 16}{59049 \cdot x^8} \] As \( x \to \infty \)[/tex], [tex]\(256/x^2\)[/tex] approaches 0, and the expression simplifies to:
\]
[tex]\( \frac{0 - 16}{59049 \cdot x^8} \rightarrow \frac{-16}{59049 \cdot x^8} \] Since \(x^8\)[/tex] in the denominator grows very large, the entire fraction tends towards 0.

Therefore, we conclude:

[tex]\[ \lim_{x \rightarrow \infty} \frac{(3-1)^8-(4 x-y)^2}{(3 x+1)^{10}} = 0 \][/tex]