According to Euler's identity
$$e^{ix}=\cos{x}+i\sin{x}$$
Using this identity
$$0^i=e^{i\ln{0}}=\cos\ln0+i\sin\ln0=\cos{-\infty}+i\sin{-\infty},\because \ln0=-\infty$$
But $\cos{-\infty}$ and $\sin{-\infty}$ are indeterminate Does it mean that $0^i$ is indeterminate.
Please Explain.
On
$Lim_{x \to 0}x^i=Lim_{x \to 0}e^{i\ln x}=Lim_{x \to 0}\cos\ln x+i\sin\ln x$
$\because$ both $\cos{\ln{0}}$ and $\sin{\ln{0}}$ are indeterminate $0^i$ is indeterminate.
On
The critical point comes from the trigonometric functions. Instead, start from the equality $$ e^{\ln x} = x $$ which holds for every positive real $x$. Then $\lim_{x \to 0^+} x = 0$, $\lim_{x \to 0^+} = -\infty$ and $\lim_{x \to 0^+} e^{\ln x} = 0$, since $x \to e^x$ is continuous on $\mathbb{R}$ and tends to $0$ when $x$ tends to $-\infty$. Thus in this case, $e^{-\infty} = 0$ is an acceptable convention, which is actually frequently used in mathematics.
However, in your case, neither $\cos x$ nor $\sin x$ have a limit when $x$ tends to $-\infty$. I hope this explains the difference between the two cases.
On
Perhaps you're confused because you expect that $0^x=0$ for any $x$. This is not true.
$0^x=0$ for any positive Real number $x$, but $0^x=\infty$ (or undefined) if $x$ is negative.
More generally, with Complex exponents,
$$b^{x+iy} = \exp\big((x+iy)\ln(b)\big)$$
$$= \exp\big(x\ln(b)+iy\ln(b)\big)$$
$$= \exp\big(x\ln(b)\big)\cdot\exp\big(iy\ln(b)\big)$$
$$= b^x\cdot\big(\cos(y\ln b)+i\sin(y\ln b)\big)$$
If we take a limit as $b\to0$, the left factor $b^x$ has the same behaviour as described above. The right factor (assuming that $y\neq0$) oscillates around the unit circle at increasing speed, because $\ln(b)\to{^-}\infty$.
If $x$ is positive, then $b^x$ damps out the oscillation, and $b^{x+iy}$ converges to $0$.
If $x$ is negative, then $b^x$ amplifies the oscillation, so it diverges.
If $x$ is exactly $0$, then $b^x=1$, and it still diverges (but not to infinity; it just spins around).
So a power of $0$ should be $0$ if the exponent's Real part is positive, $\infty$ (or undefined) if the Real part is negative, and indeterminate (or undefined) if the Real part is $0$.
(Exception: $0^0=1!$)
As you point out, one can not define $0^i$ via the identity of Euler. The question is whether we can find any other good properties of $0^i$. For me good properties would be that the function $f:x\mapsto x^i$ is continuous. This function is basically defined for $\mathbb{C}\backslash\mathbb{R-}$, where the $log function is not defined.
Can we extend the definition of the function to $0$? For this we see $$\lim_{x\in\mathbb{R},x\rightarrow 0} x^i =$$ $$=\lim_{x\in\mathbb{R},x\rightarrow 0} e^{i\cdot log(x)}$$
This limit does not converge, so we do not have any continuous extension to $0$