Left-hand limit from a table of values

For $x$ slightly less than $1$, we use the branch $h(x) = x^2$. Computing values: $h(0.9) = 0.81$, $h(0.99) = 0.9801$, $h(0.999) = 0.998001$, and $h(0.9999) = 0.99980001$. These outputs are clearly getting closer and closer to $1$ as $x$ gets closer to $1$ from below.

Since the outputs approach $1$ as $x$ approaches $1$ from the left, we conclude $\lim_{x \to 1^-} h(x) = 1$. Note this is true even though $h(1)$ itself is computed using the other branch — the limit only depends on values near $1$, not at $1$.

For $x$ slightly greater than $1$, we use the branch $h(x) = 2x - 1$. Computing values: $h(1.1) = 1.2$, $h(1.01) = 1.02$, $h(1.001) = 1.002$, and $h(1.0001) = 1.0002$. These outputs steadily decrease toward $1$ as $x$ approaches $1$ from above.

Since the outputs approach $1$ as $x \to 1^+$, we conclude $\lim_{x \to 1^+} h(x) = 1$.

From part (a), $\lim_{x \to 1^-} h(x) = 1$, and from part (b), $\lim_{x \to 1^+} h(x) = 1$. Since both one-sided limits exist and are equal to the same value, the two-sided limit exists.

Therefore $\lim_{x \to 1} h(x) = 1$. Note that this value happens to match $h(1) = 2(1)-1 = 1$ as well, but that agreement is not required for the limit to exist — only the one-sided limits needed to match each other.