$$\lim\_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}$$
In the book I am reading, the limit evaluated in this way:
$$\lim\_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}\times \frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}=\lim\_{x\to 0^+}\frac{2\tan x(1-\cos x)}{x^2+x}\times\left(\sqrt{x^2+x+1}+1\right)$$

Then it used equivalence and wrote:

$$\lim\_{x\to 0^+}\frac{2x\left(\frac12x^2\right)}{x^2+x}\times\left(\sqrt{x^2+x+1}\right)=\lim\_{x\to 0^+}\frac{x^3}{x(x+1)}.\left(\sqrt{x^2+x+1}\right)=0$$

I wonder why we should do all these calculation? Is it possible to use L'hopital rule and get $\frac0{\tfrac12}=0$ ?