Very indeterminate form: $\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x \longrightarrow (\infty-\infty)^{\infty}$ Here is problem:

$$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$

The solution I presented in the picture below was made by a Mathematics Teacher

I tried to solve this Limit without using derivative (L'hospital) and Big O notation. Although I get the answer, I don't know if the technique I'm using definitely correct.
And here is my method:
$$\begin{align*}\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x&=\lim_{x \to \infty} \left(\frac {2x}{\sqrt{x^2+2x+3} +\sqrt{x^2+3}}\right)^x\\&=\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}\end{align*}$$
Then, I define a new function here
$$y(x)=\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1$$
We have
$$\begin{align*}
\lim _{x\to\infty} y(x)&=\lim_{x \to \infty}\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1\\
&=\lim_{x \to \infty}(\sqrt{x^2+2x+3}-(x+1))+(\sqrt{x^2+3}-x)\\
&=\lim_{x \to \infty}\frac{2}{\sqrt{x^2+2x+3}+x+1}+ \lim_{x \to \infty}\frac{3}{\sqrt{x^2+3}+x}\\
&=0. 
\end{align*}$$
This implies that
$$\lim_{x \to \infty}\frac{2x}{y(x)+1}=\infty $$
Therefore,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}&=\lim_{x \to\infty} \frac{1}{ \left(\frac{y(x)+2x+1}{2x} \right)^x}\\
&=\lim_{x \to\infty} \frac{1}{ \left(1+\frac{y(x)+1}{2x} \right)^x}\\
&=\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}\\
&
\end{align*}$$
Here, we define two functions:
$$f(x)=\left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}},\quad
g(x)=\frac{y(x)+1}{2}.
$$
We deduce that,
$$
\lim_{x\to\infty} f(x)=e>0,\quad \lim_{x\to\infty} g(x)=\frac 12>0.
$$
Thus, the limit $\lim_{x\to\infty} f(x)^{g(x)} $ exists and is finite.

Finally we get,
$$\begin{align*} 
\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}
&=\frac{1}{\lim_{x \to \infty}\left( \left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}\right)}\\
&=\frac{1}{\left(\lim_{x\to\infty} \left( 1+\frac{1}{\frac{2x}{y(x)+1}} \right)^{\frac{2x}{y(x)+1}}\right)^{ \lim_{x\to\infty} \frac{y(x)+1}{2}}}\\ 
&=\frac {1}{e^{\frac12}}=\frac{\sqrt e}{e}.\\&&
\end{align*}$$

Is the method I use correct?
I have received criticisms against my work. What can I do to make the method I use, rigorous? What are the points I missed in the method?
Thank you!

By squaring, we can verify that
$$
x\le\sqrt{x^2+3}\le x\left(1+\frac3{2x^2}\right)\tag1
$$
and
$$
x+1\le\sqrt{x^2+2x+3}\le(x+1)\left(1+\frac1{x(x+1)}\right)\tag2
$$
Adding $(1)$ and $(2)$ gives
$$
2x+1\le\sqrt{x^2+2x+3}+\sqrt{x^2+3}\le(2x+1)\left(1+\frac3{2x^2}\right)\tag3
$$
Multiplying numerator and denominator by $\sqrt{x^2+2x+3}+\sqrt{x^2+3}$ gives
$$
\sqrt{x^2+2x+3}-\sqrt{x^2+3}=\frac{2x}{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}\tag4
$$
Bernoulli and cross multiplying yield
$$
1-\frac3{2x}\le\left(1-\frac3{2x^2}\right)^x\le\left(1+\frac3{2x^2}\right)^{-x}\tag5
$$
Therefore $(3)$, $(4)$, and $(5)$ yield
$$
\left(\frac{2x}{2x+1}\right)^x\left(1-\frac3{2x}\right)\le\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x\le\left(\frac{2x}{2x+1}\right)^x\tag6
$$
The Squeeze Theorem then says
$$
\lim_{x\to\infty}\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x=e^{-1/2}\tag7
$$
