solve-lim-x-2x-3-x-2-2-2x-3-4x-2-1-x-

Question Number 112974 by bemath last updated on 10/Sep/20

solve \lim_{x \to \infty} {(\frac{{2 x}^{3} - x^{2} + 2}{{2 x}^{3} - {4 x}^{2} + 1})}^{x}

Answered by john santu last updated on 10/Sep/20

Answered by mathmax by abdo last updated on 11/Sep/20

let f (x) = {(\frac{{2 x}^{3} - x^{2} + 2}{{2 x}^{3} - {4 x}^{2} + 1})}^{x} \Rightarrow f (x) = {(\frac{1 - \frac{1}{2 x} + \frac{1}{x^{3}}}{1 - \frac{2}{x} + \frac{1}{{2 x}^{3}}})}^{x}

= e^{xln (\frac{1 - \frac{1}{2 x} + \frac{1}{x^{3}}}{1 - \frac{2}{x} + \frac{1}{{2 x}^{3}}})} we do the changement \frac{1}{x} = t \Rightarrow

f (x) = f (\frac{1}{t}) = e^{\frac{1}{t} \ln (\frac{1 - \frac{t}{2} + t^{3}}{1 - 2 t + \frac{1}{2} t^{3}})} (t \to 0)

= e^{\frac{1}{t} \ln (\frac{1 - 2 t + \frac{t^{3}}{2} - \frac{t}{2} + t^{3} + 2 t - \frac{t^{3}}{2}}{1 - 2 t + \frac{t^{3}}{2}})} = e^{\frac{1}{t} \ln (1 + \frac{\frac{t^{3}}{2} + \frac{3}{2} t}{1 - 2 t + \frac{t^{3}}{2}})}

\sim e^{\frac{1}{t} (\frac{\frac{t^{3}}{2} + \frac{3}{2} t}{1 - 2 t + \frac{t^{3}}{2}})} = e^{\frac{t^{2} + 3}{2 - 4 t + t^{3}}} \to e^{\frac{3}{2}} (t \to 0) \Rightarrow \lim_{x \to \infty} f (x) = e \sqrt{e}