real analysis - Why : $1^{+infty}$ is not $1 $ however $1^{+infty}=lim _{xto 0+}(frac{sin x}{x})^{1/x}=1$?

It is well known that :$1^{+\infty}$ is indeterminate case ,I have accrossed the following problem which let me to say that :$1^{+\infty}=1$ .

$1^{+\infty}$ can be written as : $1^{+\infty}=\lim _{x\to 0+}(\frac{\sin x}{x})^{1/x}$ which is $ 1$ ,then $1^{+\infty}=1$ and it's not I.case , i don't know where i'm wrong !!!! ? and wolfram alpha says that :$\lim _{x\to 0+}(\frac{\sin x}{x})^{1/x}=1$ which mixed me .

Edit: I have edited the question to show what's mixed me in the side of

limit calculation and i don't changed my question

Answer

If you think that
$$\lim _{x\to 0^+}\bigg(\frac{\sin x}{x}\bigg)^{1/x} = \bigg(\lim_{x \to 0^+}\frac{\sin x}{x}\bigg)^{\lim_{x \to 0^+}1/x}$$
that is not correct. Here, you can find your answer I think: Why is $1^{\infty}$ considered to be an indeterminate form