\(sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0\)
We can replace sin x with x anywhere in the limit as long as x approaches 0.
Also,
\(\large \lim_{ x \to 0 } ~ x^x = 1\)
I will make the assumption that log(x)=ln(x).
The limit result can be proven if the base of log(x) is 10.
\(\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~
ormalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}\)
We get the indeterminate form 0/0, so we have to use Lhopitals rule
\(\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1\)
Therefore,
\(\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}\)
