x ln(1 + 1/x) at infinity

1. Put $u=\frac1x$. When $x\to+\infty$, $u\to0^+$ and
$$x\ln\left(1+\frac1x\right)=\frac{\ln(1+u)}{u}\longrightarrow1,$$
because $\dfrac{\ln(1+u)}u$ is the quotient $\dfrac{\ln(1+u)-\ln1}{u}$, which tends to the derivative of $\ln$ at $1$, equal to $1$.

2. For $x>0$,
$$\left(1+\frac1x\right)^x=\exp\!\left(x\ln\left(1+\frac1x\right)\right).$$
The exponent tends to $1$ and $\exp$ is continuous, so the expression tends to $e^1=e$.

Remark. The same method gives $\left(1+\frac ax\right)^x\to e^a$ for any constant $a$.