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

Your math looks good! I'd maybe just an extra step here and there to make it clear what your doing. Things like showing that you're multiplying by conjugates and maybe a change of variables, say $$z = \frac{2x}{y(x)+1},$$ near the end so it's a bit clearer where the $e$ comes from. Otherwise everything looks good! This is a tricky limit, I really like your solution.

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 $$ e^{-1/2} $$

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 $.

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 $$ \lim_{x\to\infty}\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x=e^{-1/2}\tag7 $$