limits - Solve $ lim_{xto 0} (sqrt {2x+1} - sqrt[3]{1-3x})^{x}$ without using L'Hospital

I need to solve $$ \lim_{x\to 0}\ (\sqrt {2x+1}\ -\ \sqrt[3]{1-3x})^{x}$$ Please note that I'm first year student and that this can be solved much simpler than in the answers. I tried doing $$\lim_{x\to 0} \ e^{x \cdot \ln\Bigl(\sqrt{2x+1}-1-\left(\sqrt[3]{1-3x}-1\right)\Bigr)}$$ then going with the limit inside the function like this

$$\exp\left\{\lim_{x\to0}x \cdot
\ln\left[\lim_{x \to 0}\Bigl(\sqrt{2x+1}-1\Bigr) \cdot

\lim_{x \to 0} \left(1-
\frac{ \sqrt[3]{1-3x}-1\over x }{ \sqrt{2x+1}-1 \over x }\right)\right] \right\}$$

But problem is that although I can solve third limit this way, I get that second limit is 0, which makes that 0 is inside of $\ln$ and thus is incorrect attempt. Please help, I'm new here, I wan't to contribute back and this is from my university math exam.

Answer

We shall only try to find $\lim_{x\to 0^+}$ because for negative $x$ near $0$, the power is not defined (for reasons given below).

Working without little-o stuff:

Note that $$(1+x)^2=1+2x+x^2\ge 1+2x$$ for all $x$

and of course $1<1+2x$ for all $x>0$.
We conclude that
$$1 <\sqrt{1+2x}\le 1+x\qquad\text{for }x>0.$$

Similarly,
$$ (1-x)^3=1-3x+3x^2-x^3>1-3x\qquad \text{for }x<3$$
and
$$(1-2x)^3=1-6x+12x^2-8x^3<1-3x-3x(1-4x)<1-3x \qquad \text{for }0
hence
$$1-2x<\sqrt[3]{1-3x}<1-x\qquad\text{for }0

and so
$$ x<\sqrt{1+2x}-\sqrt[3]{1-3x}<4x\qquad\text{for }0
(One can find similar bounds for negative $x$, showing that $\sqrt{1+2x}-\sqrt[3]{1-3x}\sim x<0$, and therefore $(\sqrt{1+2x}-\sqrt[3]{1-3x})^x$ is undefined for negative $x$ near $0$)

If we already know that $\lim_{x\to 0^+}x^x=1$, it follows that $(\sqrt{1+2x}-\sqrt[3]{1-3x})^x$ is squeezed between $x^x$ and $4^x\cdot x^x$ and, as $\lim_{x\to 0^+}4^x=1$, therefore also
$$ \lim_{x\to0^+}(\sqrt{1+2x}-\sqrt[3]{1-3x})^x=\lim_{x\to0^+}x^x=1.$$

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Why is $\lim_{x\to 0^+}x^x=1$?

Perhaps the most important inequality about the exponential is
$$ e^t\ge 1+t\qquad \text{for all }t\in\Bbb R.$$
Therefore, for $t>0$,
$$ e^t=(e^{t/2})^2\ge(1+\tfrac t2)^2=1+t+\frac14t^2>\frac14t^2.$$
It follows that
$$0\le \lim_{t\to +\infty}\frac{t}{e^t}\le \lim_{t\to +\infty}\frac{t}{\frac14t^2}=0.$$
With $x=e^{-t}$ (i.e., $t=-\ln x$), this becomes
$$\lim_{x\to 0^+} x\ln x=0$$
and therefore

$$\lim_{x\to0^+} x^x=\lim_{x\to 0^+} e^{x\ln x}=e^{\lim_{x\to 0^+} x\ln x} =e^0=1.$$