# What is the limit as x approaches infinity of ((2x-3)/(2x+5))^(2x+1)?

Dec 12, 2014

${\lim}_{x \to \infty} {\left(\frac{2 x - 3}{2 x + 5}\right)}^{2 x + 1}$

by the inverse property ${e}^{\setminus \ln x} = x$,

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

by the log property $\ln {x}^{r} = r \ln x$,

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

by $2 x - 3 = \frac{1}{\frac{1}{2 x - 3}}$,

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

by moving the limit inside the exponential function,

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

by l'H$\hat{\text{o}}$pital's Rule (0/0),

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

by cleaning up a bit,

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

by dividing the numerator and the denominator by ${x}^{2}$,

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

by $\frac{1}{x} \to 0$ as $x \to \infty$,

$= {e}^{- 8 \cdot {2}^{2} / \left\{2 \cdot 2\right\}} = {e}^{- 8}$

I hope that this was helpful.