To solve the limit [tex]\(\lim _{x \rightarrow \infty} \frac{(3-1)^8-(4 x-1)^2}{(3 x+1)^{10}}\)[/tex], let's break it down step by step.
1. Simplify the Numerator:
First, simplify the expression inside the numerator:
[tex]\[
(3-1)^8 = 2^8 = 256
\][/tex]
Then, express the numerator:
[tex]\[
256 - (4x - 1)^2
\][/tex]
Now expand [tex]\((4x - 1)^2\)[/tex]:
[tex]\[
(4x - 1)^2 = 16x^2 - 8x + 1
\][/tex]
So the numerator becomes:
[tex]\[
256 - (16x^2 - 8x + 1) = 256 - 16x^2 + 8x - 1 = 255 - 16x^2 + 8x
\][/tex]
Thus, the original limit becomes:
[tex]\[
\lim_{x \to \infty} \frac{255 - 16x^2 + 8x}{(3x + 1)^{10}}
\][/tex]
2. Analyze the Denominator:
The denominator is:
[tex]\[
(3x + 1)^{10}
\][/tex]
For large values of [tex]\(x\)[/tex]:
[tex]\[
(3x + 1)^{10} \approx (3x)^{10} = 3^{10} x^{10}
\][/tex]
3. Combine and Simplify:
Now combine the simplified numerator and denominator:
[tex]\[
\lim_{x \to \infty} \frac{255 - 16x^2 + 8x}{3^{10} x^{10}}
\][/tex]
Notice that the polynomial in the numerator has a lower power of [tex]\(x\)[/tex] ([tex]\(x^2\)[/tex]) compared to the polynomial in the denominator ([tex]\(x^{10}\)[/tex]).
We can factor [tex]\(x^2\)[/tex] from the numerator to better see the relationship:
[tex]\[
\lim_{x \to \infty} \frac{x^2 \left(\frac{255}{x^2} - 16 + \frac{8}{x}\right)}{3^{10} x^{10}}
\][/tex]
Simplify further:
[tex]\[
\lim_{x \to \infty} \frac{\frac{255}{x^2} - 16 + \frac{8}{x}}{3^{10} x^8}
\][/tex]
As [tex]\(x\)[/tex] approaches infinity:
[tex]\[
\frac{255}{x^2} \rightarrow 0, \quad \frac{8}{x} \rightarrow 0
\][/tex]
The expression reduces to:
[tex]\[
\lim_{x \to \infty} \frac{-16}{3^{10} x^8}
\][/tex]
Since [tex]\(x^8\)[/tex] grows very rapidly as [tex]\(x\)[/tex] approaches infinity, the overall fraction tends to 0.
4. Conclusion:
Therefore, the limit is:
[tex]\[
\lim _{x \rightarrow \infty} \frac{(3-1)^8-(4 x-1)^2}{(3 x+1)^{10}} = 0
\][/tex]
