$$\lim_{x\to 0}\cosec x \left( \sqrt{2\cos^2 x+3\cos x} - \sqrt{\cos^2 x+\sin x+4} \right)$$ is:

Rationalize the numerator.

The expression is in the form $$\frac{\infty - \infty}{0}$$. Let's rewrite $$\csc x$$ as $$1/\sin x$$ and multiply by the conjugate:

$$\lim_{x \to 0} \frac{(2 \cos^2 x + 3 \cos x) - (\cos^2 x + \sin x + 4)}{\sin x (\sqrt{2 \cos^2 x + 3 \cos x} + \sqrt{\cos^2 x + \sin x + 4})}$$

Simplify the numerator.

Numerator $$= \cos^2 x + 3 \cos x - \sin x - 4$$.

As $$x \to 0$$, we use Taylor expansions or simple substitution for the denominator's square root part: $$\sqrt{2+3} + \sqrt{1+0+4} = 2\sqrt{5}$$.

Now we deal with the limit: $$\lim_{x \to 0} \frac{\cos^2 x + 3 \cos x - \sin x - 4}{2\sqrt{5} \sin x}$$.

Using L'Hôpital's Rule on the $$0/0$$ part:

$$\frac{d}{dx}(\cos^2 x + 3 \cos x - \sin x - 4) = -2\cos x \sin x - 3\sin x - \cos x$$

At $$x=0$$, this equals $$0 - 0 - 1 = -1$$.

The derivative of the denominator part ($$\sin x$$) is $$\cos x$$, which is $$1$$ at $$x=0$$.

Result $$= \frac{-1}{2\sqrt{5}}$$.