Answer
$a=\frac{1}{2}$
Work Step by Step
Use the decomposition it follows that the given limit can be written as:
$$\lim_{x \to \infty}(1+\frac{2a}{x-a})^x$$
Put $x-a=t$ so when $x$ goes to $\infty$ it follows that $t$ goes to $\infty$ too so:
$$\lim_{x \to \infty}(1+\frac{2a}{x-a})^x=\lim_{t \to \infty}(1+\frac{2a}{t})^{t+a}=\lim_{t \to \infty}(1+\frac{2a}{t})^{t}\cdot (1+\frac{2a}{t})^{a}=\lim_{t \to \infty}(1+\frac{2a}{t})^{t} \cdot \lim_{t \to \infty} (1+\frac{2a}{t})^{a}=\lim_{t \to \infty}(1+\frac{2a}{t})^{t} \cdot 1=\lim_{t \to \infty}(1+\frac{2a}{t})^{t} $$
so:
$$\lim_{t \to \infty}(1+\frac{2a}{t})^{t}=e $$
The above equation is true if:
$$2a=1$$
$$a=\frac{1}{2}$$
