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!

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:
$$\lim_{x\rightarrow+\infty}(\sqrt{x^2+2x+4}-x)^x =\lim_{x\rightarrow+\infty}\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}\right)^x =$$
$$=\lim_{x\rightarrow+\infty}\left(1+\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)^{\frac{1}{\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1}\cdot x\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)} =$$
$$=e^{\lim\limits_{x\rightarrow+\infty}x\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)}=e^{\lim\limits_{x\rightarrow+\infty}\frac{x( x+4-\sqrt{x^2+2x+4})}{\sqrt{x^2+2x+4}+x}}=$$
$$=e^{\lim\limits_{x\rightarrow+\infty}\frac{x(6x+12)}{\left(\sqrt{x^2+2x+4}+x\right)\left( x+4+\sqrt{x^2+2x+4}\right)}}=e^{\frac{6}{2\cdot2}}=e^{\frac{3}{2}}.$$
