$$\lim_{x \to 0} \frac{x + 2\sin x}{\sqrt{x^2 + 2\sin x + 1} - \sqrt{\sin^2 x - x + 1}}$$ is

We begin with the limit

$$L=\lim_{x\to0}\dfrac{x+2\sin x}{\sqrt{x^{2}+2\sin x+1}-\sqrt{\sin^{2}x-x+1}}.$$

Both the numerator and the denominator approach $$0$$ as $$x\to0$$, so L’Hôpital’s Rule is applicable. L’Hôpital’s Rule states that for functions $$f(x)$$ and $$g(x)$$ which satisfy $$f(0)=g(0)=0$$ and have derivatives continuous near $$0$$, we have

$$\lim_{x\to0}\dfrac{f(x)}{g(x)}=\lim_{x\to0}\dfrac{f'(x)}{g'(x)}$$

provided the limit on the right exists.

We set

$$f(x)=x+2\sin x,\qquad g(x)=\sqrt{x^{2}+2\sin x+1}-\sqrt{\sin^{2}x-x+1}.$$

Derivative of the numerator

$$f'(x)=1+2\cos x.$$

Therefore

$$f'(0)=1+2\cos0=1+2\cdot1=3.$$

Derivative of the denominator

Write $$g(x)=u(x)-v(x)$$ with

$$u(x)=\sqrt{x^{2}+2\sin x+1},\qquad v(x)=\sqrt{\sin^{2}x-x+1}.$$

Using $$\dfrac{d}{dx}\sqrt{h(x)}=\dfrac{h'(x)}{2\sqrt{h(x)}}$$ we find

$$u'(x)=\dfrac{2x+2\cos x}{2\sqrt{x^{2}+2\sin x+1}} =\dfrac{x+\cos x}{\sqrt{x^{2}+2\sin x+1}},$$

$$v'(x)=\dfrac{2\sin x\cos x-1}{2\sqrt{\sin^{2}x-x+1}} =\dfrac{\sin x\cos x-\dfrac12}{\sqrt{\sin^{2}x-x+1}}.$$

Hence

$$g'(x)=u'(x)-v'(x) =\dfrac{x+\cos x}{\sqrt{x^{2}+2\sin x+1}} -\dfrac{\sin x\cos x-\dfrac12}{\sqrt{\sin^{2}x-x+1}}.$$

Now evaluate $$g'(x)$$ at $$x=0$$. We know $$\sin0=0,\;\cos0=1$$, and both radicals become $$\sqrt1=1$$. Substituting,

$$g'(0)=\dfrac{0+1}{1}-\dfrac{0\cdot1-\dfrac12}{1} =1-\Bigl(-\dfrac12\Bigr)=1+\dfrac12=\dfrac32.$$

Applying L’Hôpital’s Rule

$$L=\lim_{x\to0}\dfrac{f(x)}{g(x)} =\lim_{x\to0}\dfrac{f'(x)}{g'(x)} =\dfrac{f'(0)}{g'(0)} =\dfrac{3}{\dfrac32}=2.$$

So, the answer is $$2$$.