The answer should be ✓3/6

Your proposed answer $\frac{\sqrt{3}}{6}$ is incorrect.

After simplifying the numerator, we had $2x$, not $x$.

$$\lim_{x \to 0} \frac{2x}{x(\sqrt{2x+3} + \sqrt{3})}$$

Canceling $x$ gives:

$$\lim_{x \to 0} \frac{2}{\sqrt{2x+3} + \sqrt{3}}$$

Substituting $x=0$:

$$\frac{2}{\sqrt{2(0)+3} + \sqrt{3}} = \frac{2}{\sqrt{3} + \sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$