Checking the three-part continuity test at x = 0

First check that $a(0)$ is defined: by the piecewise definition, $a(0) = \frac{1}{2}$, so condition 1 is satisfied. Next compute the limit as $x \to 0$ using the formula for $x \neq 0$: $\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \lim_{x\to 0}\frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{x(\sqrt{x+1}+1)} = \lim_{x\to 0}\frac{(x+1)-1}{x(\sqrt{x+1}+1)}.$

Simplifying, this becomes $\lim_{x\to 0}\frac{x}{x(\sqrt{x+1}+1)} = \lim_{x\to 0}\frac{1}{\sqrt{x+1}+1} = \frac{1}{\sqrt{1}+1} = \frac{1}{2}.$ So the limit exists and equals $\frac{1}{2}$, satisfying condition 2. Since this limit equals $a(0) = \frac{1}{2}$, condition 3 also holds, so $a(x)$ is continuous at $x=0$.

No, it would not stay continuous. The limit as x approaches zero was found to equal one half, but the function value would now be one. Since the limit no longer equals the function value at that point, condition three would fail, making the function discontinuous at zero.