Limit Question: doubting about my answer! A limit where $x \to \infty$ Hey guys so I have this limit:
$$\lim_{x \to ∞} f(x) = \frac{(3x+1)^3(x-1)}{(x-2)^4}$$
and I got $27$ as final answer, just wondering if you guys can check;
I expanded the numerator and denominator and then divided everything by $x^4$

Note that
$$\frac{(3x+1)^3(x-1)}{(x-2)^4}=27\frac{(x+1/3)^3(x-1)}{(x-2)^4}=27\frac{\frac{(x+1/3)^3}{x^3}\frac{(x-1)}x}{\frac{(x-2)^4}{x^4}}\\=27\frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}\to27\frac{(1+0)^3(1-0)}{(1-0)^4}=27$$
as $x\to\infty$, so your result is correct.
