Euler

Note first that $\dfrac{2x + 1}{2x - 1} = 1 + \dfrac{2}{2x - 1}$ . The limit can then be written as

$L = \displaystyle\lim_{x \to \infty} \left( \left(1 + \dfrac{2}{2x - 1}\right)\left(1 + \dfrac{2}{2x - 1}\right)^{2x - 1}\right) =$

$\displaystyle\lim_{y \to \infty} \left(1 + \dfrac{2}{y}\right) \times \lim_{y \to \infty} \left(1 + \dfrac{2}{y}\right)^{y}$ ,

where $2x - 1 = y \to \infty$ as $x \to \infty$ . The first of these limits goes to $1$ , and the second to $e^{2}$ , so

$L = e^{2} = \boxed{7.389}$ to 3 decimal places.

To see how the second limit comes about, let $t = \dfrac{y}{2}$ . then $t \to \infty$ as $y \to \infty$ , making the limit

$\displaystyle\lim_{t \to \infty} \left(1 + \dfrac{1}{t}\right)^{2t} = \left(\lim_{t \to \infty} \left(1 + \dfrac{1}{t}\right)^{t}\right)^{2} = e^{2}$ , since by definition $\displaystyle\lim_{t \to \infty} \left(1 + \dfrac{1}{t}\right)^{t} = e.$

Let $f(x)=\dfrac{2x+1}{2x-1}$ and $g(x)=2x$ (easier for me to type :P).

Since, $\displaystyle \lim_{x \to \infty}{f(x)=1}$ and $\displaystyle \lim_{x \to \infty}{(f(x)-1)g(x)}$ exists(it is equal to 2),

$\displaystyle \lim_{x \to \infty}{f(x)^{g(x)}}=e^{\lim_{x \to \infty}{(f(x)-1)g(x)}}=e^2 \approx 7.389$ .