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!
 A: This is evaluated with  Taylor's formula at order $2$:
$$\ln\Bigl(\bigl(\sqrt{x^2+2x+4}-x\bigr)^x\Bigr)=x\ln\bigl(\sqrt{x^2+2x+4}-x\bigr)=x\Bigl(\ln x+\ln\biggl(\sqrt{1+\frac2x+\frac4{x^2}}-1\biggr)\biggr).
$$
Now, Taylor's expansion at order $2$ yields
\begin{align}\sqrt{1+\frac2x+\frac4{x^2}}-1&=1+\frac12\biggl(\frac2x+\frac4{x^2}\biggr)-\frac18\biggl(\frac2x+\frac4{x^2}\biggr)^2+o\biggl(\frac1{x^2}\biggr)-1\\
&=\frac1x+\frac3{2x^2}+o\biggl(\frac1{x^2}\biggr)=\frac1x\biggl(1+\frac3{2x}+o\biggl(\frac1{x}\biggr)\biggr),\end{align}
so that
$$\ln\biggl(\sqrt{1+\frac2x+\frac4{x^2}}-1\biggr)=-\ln x+\ln\biggl(1+\frac3{2x}+o\biggl(\frac1{x}\biggr)\biggr)=-\ln x+\frac3{2x}+o\biggl(\frac1{x}\biggr),$$
and finally
$$\ln\Bigl(\bigl(\sqrt{x^2+2x+4}-x\bigr)^x\Bigr)=x\biggl(\ln x-\ln x+\frac3{2x}+o\biggl(\frac1{x}\biggr)\biggr)=\frac32+o(1)\to\frac32.$$
A: Sometimes WA gives  absolutely wrong results, but in our case he gave a right answer.
Since $f(x)=e^x$ is a continuous function, we obtain:
$$\lim_{x\rightarrow+\infty}(\sqrt{x^2+2x+4}-x)^x =\lim_{x\rightarrow+\infty}\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}\right)^x =$$
$$=\lim_{x\rightarrow+\infty}\left(1+\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)^{\frac{1}{\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1}\cdot x\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)} =$$
$$=e^{\lim\limits_{x\rightarrow+\infty}x\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)}=e^{\lim\limits_{x\rightarrow+\infty}\frac{x( x+4-\sqrt{x^2+2x+4})}{\sqrt{x^2+2x+4}+x}}=$$
$$=e^{\lim\limits_{x\rightarrow+\infty}\frac{x(6x+12)}{\left(\sqrt{x^2+2x+4}+x\right)\left( x+4+\sqrt{x^2+2x+4}\right)}}=e^{\frac{6}{2\cdot2}}=e^{\frac{3}{2}}.$$
A: 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}
$.
