<s>[INST] 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!
 [/INST] The solution appears to be correct. Just for the sake of sanity, here's a different argument, based on the idea that knowing derivatives is knowing many limits.
First, find the limit of the logarithm of the beast, which is best treated also with the substitution $x=1/t$, which makes us trying to find
$$
\lim_{t\to0^+}\frac{1}{t}\log\left(\frac{\sqrt{1+2t+3t^2}-\sqrt{1+3t^2}}{t}\right)
=
\lim_{t\to0^+}\frac{1}{t}\log\left(\frac{2}{\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}}\right)
$$
This can be rewritten as
$$
\lim_{t\to0^+}-\frac{\log\bigl(\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}\,\bigr)-\log2}{t}
$$
which is the negative of the derivative at $0$ of
$$
f(t)=\log\bigl(\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}\,\bigr)
$$
Since
$$
f'(t)=\frac{1}{\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}}\left(\frac{1+3t}{\sqrt{1+2t+3t^2}}+\frac{3t}{\sqrt{1+3t^2}}\right)
$$
we have $f'(0)=1/2$ and therefore the limit is $-1/2$, so your given limit
$$
e^{-1/2}
$$
</s>