Question

Evaluate:${lim}_{x\to \infty }{\left(\frac{2x^2+3}{2x^2+5}\right)}^{8x^2+3}$

Answer 1

$=e^{{lim}_{x\to \infty }(\frac{2x^2+3}{2x^2+5}-1)(8x^2+3)}$

$=e^{{lim}_{x\to \infty }\left(\frac{2x^2+3-2x^2-5}{2x^2+5}\right)(8x^2+3)}$

$=e^{{lim}_{x\to \infty }\left(\frac{-2(8x^2+3)}{2x^2+5}\right)}$

$=e^{{lim}_{x\to \infty }\left(\frac{-2x^2(8+\frac{3}{x^2})}{x^2(2+\frac{5}{x^2})}\right)}$

$=e^{{lim}_{x\to \infty }\left(\frac{-2(8+\frac{3}{x^2})}{(2+\frac{5}{x^2})}\right)}$

$=e^{\left(\frac{-2(8+0)}{(2+0)}\right)}$

$=e^{\left(\frac{-16}{2}\right)}$

$=e^{-8}$