<s>[INST] Wrong result for a $1^\infty$ limit I wanna know the $$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x .$$
I applied $1^\infty$ algorithm and I have in last sentence $$\lim_{x\rightarrow\infty} \left(\frac{2x+4-\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right)^x.$$ After that I used $x=e^{\ln x}$ and got this: $$\lim_{x\rightarrow\infty} \left(\frac{x^2+4x-x\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right).$$ The result for this, I think is $2$, but I'm not really sure of this, Wolfram-Alpha says that it's $\frac{3}{2}$. Can explain somebody why, but without L'Hospital? 
Thks a lot!
 [/INST] Want
$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x
$.
Since,
for small $z$,
$\sqrt{1+z}
=1+\frac12 z + O(z^2)
$
and
$\ln(1+z)
=z+O(z^2)
$,
$\begin{array}\\
\sqrt{x^2+2x+4}
&=\sqrt{(x+1)^2+3}\\
&=(x+1)\sqrt{1+\frac{3}{(x+1)^2}}\\
&=(x+1)(1+\frac{3}{2(x+1)^2}+O(x^{-4}))\\
&=x+1+\frac{3}{2(x+1)}+O(x^{-3})\\
\text{so}\\
\sqrt{x^2+2x+4}-x
&=1+\frac{3}{2(x+1)}+O(x^{-3})\\
\end{array}
$
Therefore
$\ln(\sqrt{x^2+2x+4}-x)
=\ln(1+\frac{3}{2(x+1)}+O(x^{-3}))
=\frac{3}{2(x+1)}+O(x^{-2})
$
so
$\begin{array}\\
x\ln(\sqrt{x^2+2x+4}-x)
&=\frac{3x}{2(x+1)}+O(x^{-1})\\
&=\frac{3(x+1)-3}{2(x+1)}+O(x^{-1})\\
&=\frac32-\frac{3}{2(x+1)}+O(x^{-1})\\
&=\frac32+O(x^{-1})\\
\end{array}
$
so that
$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x
=e^{3/2}
$.
</s>