Three more classical limits

a) For $x\neq0$, $-1\le\sin(1/x)\le1$, so $1\le 2+\sin(1/x)\le3$. Multiplying by $x>0$ gives $x\le x(2+\sin(1/x))\le3x$; both bounds tend to $0$, so the right limit is $0$. Multiplying by $x<0$ reverses the inequalities; again both bounds tend to $0$. The limit is $0$.

b) Write
$$\frac{\ln(1+3x)}{2x}=\frac{3}{2}\cdot\frac{\ln(1+3x)}{3x}\longrightarrow\frac{3}{2}\cdot1=\frac{3}{2}.$$

c) Write $(1+x)^{\ln x}=\exp(\ln x\cdot\ln(1+x))$. As $x\to0^+$,
$$\ln x \cdot \ln(1+x) = x \ln x \cdot \dfrac{\ln(1+x)}{x} \rightarrow 0.$$
Hence the limit is $e^0=1$.