Solving a limit without reaching an indeterminate expression Find the limit of :
$$\lim\limits_{x \to 0^+}{(2\sqrt{x}+x)^\frac{1}{\ln x}}{}$$
I've tried to make it look like an exponent of e:
$$e^\frac{\ln (2*\sqrt{x}+x)}{\ln x}$$
but, then again I reach an indeterminate form of infinity divides by infinity.
I then tried to use L'Hospital's rule, which also seems not to work.

Note $x^{1/\ln x} = e,$ which follows by applying $\ln$ to both sides. Thus $\sqrt x^{1/\ln x} = e^{1/2}.$
Now $x<\sqrt x$ for $0<x<1,$ and because the power $1/\ln x<0,$ we have
$$(3\sqrt x)^{1/\ln x} < (2\sqrt x+x)^{1/\ln x} < (2\sqrt x)^{1/\ln x}$$
for $x$ in this range. The left side equals $3^{1/\ln x}e^{1/2}.$ As $x\to 0^+,$ the limit of this is $3^0\cdot e^{1/2}= e^{1/2}.$ Similarly the right side $\to e^{1/2}.$ Therefore $e^{1/2}$ is the desired limit.
