(1 + sin x)^(1/x)

Idea. A power whose base tends to $1$ and whose exponent tends to infinity is indeterminate. Taking logarithms moves the difficulty to a product, which is easier to analyse.

Computation. For $x$ near $0$, $1+\sin x>0$, so
$$(1+\sin x)^{1/x}=\exp\!\left(\frac{\ln(1+\sin x)}{x}\right).$$
For $x\neq0$ small, $\sin x\neq0$, and we can write
$$\frac{\ln(1+\sin x)}{x}=\frac{\ln(1+\sin x)}{\sin x}\cdot\frac{\sin x}{x}.$$
As $x\to0$: $u=\sin x\to0$ so $\dfrac{\ln(1+u)}u\to1$, and $\dfrac{\sin x}x\to1$. The exponent tends to $1$, and since $\exp$ is continuous the limit is $e$.

Remark. The only subtle point is to make sure the quantity we divide by, $\sin x$, is not $0$ near $x=0$ (apart from $x=0$ itself).