Solución detallada
Tomamos como el límite
$$\lim_{x \to \infty} \left(\frac{x + 3}{x + 5}\right)^{1 - 2 x}$$
cambiamos
$$\lim_{x \to \infty} \left(\frac{x + 3}{x + 5}\right)^{1 - 2 x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 5\right) - 2}{x + 5}\right)^{1 - 2 x}$$
=
$$\lim_{x \to \infty} \left(- \frac{2}{x + 5} + \frac{x + 5}{x + 5}\right)^{1 - 2 x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{2}{x + 5}\right)^{1 - 2 x}$$
=
hacemos el cambio
$$u = \frac{x + 5}{-2}$$
entonces
$$\lim_{x \to \infty} \left(1 - \frac{2}{x + 5}\right)^{1 - 2 x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u + 11}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{11} \left(1 + \frac{1}{u}\right)^{4 u}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{11} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{4}$$
El límite
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
hay el segundo límite, es igual a e ~ 2.718281828459045
entonces
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{4} = e^{4}$$
Entonces la respuesta definitiva es:
$$\lim_{x \to \infty} \left(\frac{x + 3}{x + 5}\right)^{1 - 2 x} = e^{4}$$