Three basic limits by conjugation and standard limits

a) Direct substitution gives $0/0$. Multiply by the conjugate:
$$\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\frac{(1+x)-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})}=\frac{2}{\sqrt{1+x}+\sqrt{1-x}}.$$
As $x\to0$, the denominator tends to $\sqrt1+\sqrt1=2$, so the limit is $1$.

b) Direct substitution gives $0/0$. Simplify:
$$\frac{\frac1x-1}{x-1}=\frac{(1-x)/x}{x-1}=-\frac{1}{x}\quad(x\neq1).$$
Hence the limit as $x\to1$ is $-1$.

c) Direct substitution gives $0/0$. Write
$$\frac{\sin 2x}{\sin 3x}=\frac{2}{3}\cdot\frac{\sin 2x/(2x)}{\sin 3x/(3x)}\longrightarrow\frac{2}{3}\cdot\frac{1}{1}=\frac{2}{3}.$$