Why $\lim\limits_{x\to 0}\left(x^{2x}+2x^{1+2x}\cdot \ln x\right) = 1$?

Question

As title says - I don't see why following equation is true:

$$\lim\limits_{x\to 0}\left(x^{2x}+2x^{1+2x}\cdot \ln x\right) = 1$$

I know that

$\lim\limits_{x \to 0}x^{2x} = 0^0 = 1$,

and

$\lim\limits_{x \to 0}2x^{1+2x} = 2\cdot 0^1 = 0$,

$\lim\limits_{x \to 0}\ln{x} = -\infty$

leaves with

$0 \cdot(-\infty) = 0$, but, $0 \cdot (-\infty)$ is not defined. So how can it be $0$ ?$^*$

I don't see how to simplify the original expression to cancel terms out or do some other trickery.
Requesting for explanation.

*Edit for clarification.

Answer 1

Hint: $$\lim_{x\to 0^+}x\ln(x)=0$$

Answer 2

Because $\lim\limits_{x\rightarrow0^+}x^x=1$ and $\ln$ is a continuous function.

We obtain: $$\lim_{x\to 0}\left(x^{2x}+2x^{1+2x}\cdot \ln{x}\right) =\lim_{x\to 0}\left(\left(x^{x}\right)^2+2\left(x^{x}\right)^2\cdot \ln{x^x}\right)=1^2+2\cdot1^2\cdot\ln1=1$$

Answer 3

Let's show that $$\lim_{x\to 0^+}x\ln(x)=\lim_{x\to 0^+}\frac{\ln(x)}{1/x}=[ L’Hopital’s]=\lim_{x\to 0^+}\frac{1/x}{-1/x^2}=\lim_{x\to 0^+}(-x)=0$$

Answer 4

Let $y=x^x$. Then $\ln y=x\ln x\implies y=e^{x\ln x}=x^x.$ Thus, we have the following limit: $$ \lim\limits_{x\to 0}x^x=\lim\limits_{x\to 0}e^{x\ln x}=e^{\lim\limits_{x \to 0}(x\ln x)}=e^{0}=1.$$ Now by substituting 1 instead of $\lim\limits_{x\to 0}x^x$, you will obtain the desired result.