Limits : $x^{-1} \sin(x) $ as x -> 0 and $\tfrac{\ln(x)}{1-x}$ as x-> 0

a) Using the sine expansion formula
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5}_···$$
We get
$$\lim_{x \to 0}\frac{\sin x}{x}=\lim_{x \to 0}1-\frac{x^2}{3!}+\frac{x^4}{5}_···$$
Put the limit now we get
$$\lim_{x \to 0}\frac{\sin x}{x}=1-0+0_···=\boxed1$$

b) $$\lim_{x \to 0}\frac{\ln x}{1-x}$$
This limit is not indeterminate as it is not one of the seven indeterminate forms and it can be solved  simply by directly substituting the limit.
Note that here for $x<0$, logarithm function is not defined hence the question should be assumed as
$$\lim_{x \to 0^+}\frac{\ln x}{1-x}=\frac{\ln 0}{1-0}=\boxed{-\infty}$$