What are some other ways of solving $y=\lim_{x \to 0^+} (2\sin(\sqrt x) + \sqrt x\sin{1\over x})^x$? I have come across many limit questions of the form $0^0$
For instance , here is one example
$$y=\lim_{x \to 0^+} (2\sin(\sqrt x) + \sqrt x\sin{1\over x})^x$$
In this example if I take logarithm on both sides , it becomes
$$\ln y = \lim_{x \to 0^+} x\ln 
(2\sin(\sqrt x) + \sqrt x\sin{1\over x})$$
And then I make the argument that as $x \to 0$ ,$x$ reaches zero more dominantly than $\ln 
(2\sin(\sqrt x) + \sqrt x\sin{1\over x})$ reaches $-\infty$ . And so $\ln y = 0$ , thus , $y=1$
But I realised tht this the argument I make in almost all such cases , and it got to me thinking are all limits of the form $0^0$ eventually turn out to be 1 ?
So what I am looking for is :
(a) Various other ways of solving the above limit (as many different ways are possible) , but please avoid using L'hopital's rule as it would be too tedious to work with here.
(b) Counter examples , that is , limits of the form $0^0$ which are not equal to $1$ 
Thanks for your time !
Edit : I got the plenty counter examples , now I am only looking for other ways of solving the above limit
 A: $\lim\limits_{x \to 0} \left( 2x  \right)^{\frac{3}{5 - \ln(4x)}} = \left[ 0^0 \right] = \lim\limits_{x \to 0} e^{\frac{3 \cdot \ln(2x)}{5 - \ln(4x)}} = \left[ \text{L'Hopital}  \right] = \lim\limits_{x \to 0} e^{\frac{\frac{d}{dx}(3 \cdot \ln(2x))}{\frac{d}{dx}(5 - \ln(4x))}} = \lim\limits_{x \to 0} e^{\frac{\frac{3}{x}}{\frac{-1}{x}}} = \boxed{e^{-3}}$
A: $0^0$ we call "Uncertainty", because if we take different sequences with limits $0$ then we obtain different results:
$$\begin{array} {}
\left ( \frac{1}{n} \right )^{\frac{1}{n}}=e^{\frac{1}{n} \ln \frac{1}{n}} \to 1 \\
\left ( \frac{1}{n} \right )^{e^{-n^2}} = e^{\frac{1}{n} \ln e^{-n^2}} \to 0 \\
\left ( \frac{1}{n} \right )^{e^{-n}} = e^{\frac{1}{n} \ln e^{-n}} \to e^{-1}
\end{array}
$$
last line makes clear, that we can obtain any finite number $a>0$. One casuistic example we obtain if allow negative base (somebody forbid complex numbers?)
$$\left ( -\frac{1}{n} \right )^{e^{-n^2}} = e^{-\frac{1}{n} \ln e^{-n^2}} \to +\infty$$
and last requirement
$$\lim_{x \to 0^+} (2\sin(\sqrt x) + \sqrt x\sin{1\over x})^x = \lim_{x \to 0^+}e^{x \ln (2\sin(\sqrt x) + \sqrt x\sin{1\over x})}$$
now
$$x \ln (2\sin(\sqrt x) + \sqrt x\sin{1\over x})=x \ln(2\sin(\sqrt x))+x\ln \left (1+\frac{\sqrt x\sin{1\over x}}{2\sin(\sqrt x)} \right ) \to0$$
A: No, the limit of the form 0^0 not always comes out to be 1
Here is a counter example,
https://www.youtube.com/watch?v=89d5f8WUf1Y
A: Here is an approach which takes care of the oscillating term $\sin(1/x)$.
Clearly we have $|\sin(1/x)|\leq 1$ and therefore $$2\sin\sqrt{x}-\sqrt{x}\leq 2\sin\sqrt{x}+\sqrt {x} \sin\frac {1}{x}\leq 2\sin\sqrt{x}+\sqrt{x}$$ Raising each term to power $x$ we get an inequality of the form $g(x) \leq f(x) \leq h(x) $ where $f(x)$ is the function under limit in question. You can now show that $g(x) \to 1,h(x)\to 1$ so that by squeeze theorem $f(x) \to 1$.
We have $$g(x) =(2\sin\sqrt{x}-\sqrt{x})^x=x^{x/2}\left(2\cdot\frac{\sin\sqrt{x}}{\sqrt{x}}-1\right)^x$$ The expression $x^{x/2}$ tends to $1$ (prove this via the standard limit that $x\log x\to 0$) and the expression in large parentheses tends to $1$ so that the second factor also tends to $1$. It follows that $g(x) \to 1$ and in a similar manner $h(x) \to 1$.
A: Although some authors define $0^0=1$, does not mean that
$$
\lim_{(x,y)\to(0,0)}x^y=1\tag1
$$
There are many paths to $(0,0)$ that do not lead to $1$. For example, if $x\gt0$,
$$
x^{\frac\lambda{\log(x)}}=e^\lambda\tag2
$$
Taking the limit as $x\to0^+$ obviously gives $e^\lambda$.

However, for $0\le x\le\frac{\pi^2}{16}$, the concavity of $\sin(x)$ gives $\frac{2\sqrt2}\pi\sqrt{x}\le\sin\left(\sqrt{x}\right)$. Therefore,
$$
\left(\frac{4\sqrt2}\pi-1\right)\sqrt{x}\le2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\tfrac1x\right)\le3\sqrt{x}\tag3
$$
Furthermore, for $a\gt0$,
$$
\begin{align}
\lim_{x\to0^+}\left(a\sqrt{x}\right)^{\,x}
&=\lim_{x\to0^+}\left(\sqrt2\,a\right)^x\lim_{x\to0^+}(x/2)^{x/2}\\[3pt]
&=1\cdot1\tag4
\end{align}
$$
The Squeeze Theorem, with $(3)$ and $(4)$, shows that
$$
\lim_{x\to0^+}\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\tfrac1x\right)\right)^x=1\tag5
$$
