Let $$\displaystyle y= \lim_{x\rightarrow \infty}\left(\frac{2x}{2x+1}\right)^x\;,$$ Now $$\displaystyle \ln(y) = \lim_{x\rightarrow \infty}x\cdot \ln\left[\frac{2x}{2x+1}\right] = \lim_{x\rightarrow \infty}x\cdot \left[\ln(2x)-\ln(2x+1)\right]$$
Now $$\displaystyle \ln(y) = \lim_{x\rightarrow \infty}\frac{\left[\ln(2x)-\ln(2x+1)\right]}{\frac{1}{x}}$$
applying $\bf{L,Hopital\; Rule}$
$$\displaystyle \ln(y) = \lim_{x\rightarrow \infty}\frac{\frac{2}{2x}-\frac{2}{2x+1}}{-\frac{1}{x^2}} = -\lim_{x\rightarrow \infty}\frac{2x^2}{2x\cdot (2x+1)} = -\frac{1}{2}$$
so we get $$\displaystyle \ln(y) = -\frac{1}{2}\Rightarrow y = e^{-\frac{1}{2}}$$