Left-hand limit from a table

For $x \le 0$, $h(x) = e^x$, so compute $e^{-0.1} \approx 0.9048$, $e^{-0.01} \approx 0.9900$, and $e^{-0.001} \approx 0.9990$. As $x$ gets closer to $0$ from the left, these values are getting closer and closer to $1$.

This trend shows that as $x \to 0^-$, $h(x) \to 1$. Note that this matches $e^0 = 1$, since $e^x$ is continuous, so $\displaystyle\lim_{x \to 0^-} h(x) = 1$.

For $x > 0$, $h(x) = x + 1$, so compute $h(0.1) = 1.1$, $h(0.01) = 1.01$, and $h(0.001) = 1.001$. As $x$ approaches $0$ from the right, these values are getting closer and closer to $1$.

So $\displaystyle\lim_{x \to 0^+} h(x) = 1$.

From the previous parts, $\displaystyle\lim_{x \to 0^-} h(x) = 1$ and $\displaystyle\lim_{x \to 0^+} h(x) = 1$. Since both one-sided limits exist and are equal to the same number, the two-sided limit exists.

Therefore $\displaystyle\lim_{x \to 0} h(x) = 1$, even though this value happens also to equal $h(0) = e^0 = 1$ in this case.

No, it does not. The limit as x approaches zero depends only on the values of h near zero, not on h at zero itself. In this problem the limit happened to equal h of zero, but that was a coincidence, not a requirement. The limit is determined purely by the trend in the tables approaching from each side.