# How do you find the limit lim (3^(x+1)-2^(x+4))/(3^(x-2)+2^(x-1)+6) as x->oo?

$3$
${\lim}_{x \to \infty} \frac{{3}^{x + 1} - {2}^{x + 4}}{{3}^{x - 2} + {2}^{x - 1} + 6} = {\lim}_{x \to \infty} \frac{{3}^{-} 1 - {2}^{4} \cdot {\left(\frac{2}{3}\right)}^{x}}{{3}^{-} 2 + {2}^{-} 1 {\left(\frac{2}{3}\right)}^{x} + \frac{6}{3} ^ x} = 3$