Left-hand limit at x = 0

Since we are approaching $0$ from the left, we only consider values of $x$ slightly less than $0$, so we use the rule $f(x) = e^x$, which applies for $x < 0$.

As $x \to 0^-$, $e^x \to e^0 = 1$. Therefore $\displaystyle\lim_{x \to 0^-} f(x) = 1$.

Since we are approaching $0$ from the right, we consider values of $x$ slightly greater than $0$, so we use the rule $f(x) = x+1$, which applies for $x \ge 0$.

As $x \to 0^+$, $x+1 \to 0+1 = 1$. Therefore $\displaystyle\lim_{x \to 0^+} f(x) = 1$.

The left-hand limit is $1$ and the right-hand limit is also $1$, as found previously.

Since the one-sided limits agree, the two-sided limit exists and equals their common value: $\displaystyle\lim_{x \to 0} f(x) = 1$.

The right-hand limit was computed using the piece defined for x greater than or equal to zero.

That piece is the linear formula, x plus one, not the exponential formula.

So the correct answer is linear.