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!

Your approach is correct but its presentation / application is more complicated than needed here.
Here is how you can use the same approach with much less effort. You have already observed that the base $$F(x) =\sqrt {x^2+2x+3}-\sqrt{x^2+3}$$ tends to $1$ as $x\to\infty $. Now the expression under limit can be written as $$\{F(x) \} ^x=\{\{1+(F(x)-1)\}^{1/(F(x)-1)}\}^{x(F(x)-1)}$$ The inner expression tends to $e$ and the exponent $x(F(x) - 1)\to -1/2$ so that the desired limit is $e^{-1/2}$.

Another part of your approach is that it involves the tricky use of subtracting $2x+1$ from $y(x) $. For those who are experienced in the art of calculus this step is obvious via the approximation $$\sqrt{x^2+2ax+b}\approx x+a$$ but it may appear a bit mysterious for a novice. It is best to either explain this part or remove it altogether as I have done it in my answer.
Also note that your approach uses the following limits / rules (it is not necessary to point them out explicitly unless demanded by some strict examiner) :


*

*$\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x=e$

*If $\lim_{x\to\infty} f(x) =a>0$ and $\lim_{x\to\infty} g(x) =b$ then $\{f(x) \} ^{g(x)} \to a^b$ as $x\to\infty $.

