The function $h$ is defined piecewise as follows.
\[
h(x)=\left\{\begin{array}{ll}
-x^{2}+x+1 & \text { if } x \neq 1 \\
3 & \text { if } x=1
\end{array}\right.
\]
Find the following limits.
If a limit does not exist, click on "Does Not Exist."
(a) $\lim _{x \rightarrow 1^{+}} h(x)=$ Џ్
(b) $\lim _{x \rightarrow 1^{-}} h(x)=$
(c) $\lim _{x \rightarrow 1} h(x)=$

Step 1 ：Define the function h(x) as follows: \[h(x)=\left\{\begin{array}{ll}-x^{2}+x+1 & \text { if } x \neq 1 \\ 3 & \text { if } x=1\end{array}\right.\]

Step 2 ：Find the limit of h(x) as x approaches 1 from the right (1+). Since we are approaching 1 but not actually reaching it, we use the definition for x≠1. The limit is 1.

Step 3 ：Find the limit of h(x) as x approaches 1 from the left (1-). Again, we use the definition for x≠1. The limit is also 1.

Step 4 ：Find the limit of h(x) as x approaches 1 directly. This means we need to check if the limits from the right and left are the same. If they are, that is the limit of the function as x approaches 1. If they are not, the limit does not exist. In this case, the limits from the right and left are the same, so the limit as x approaches 1 directly exists and is also 1.

Step 5 ：Final Answer: (a) \(\lim _{x \rightarrow 1^{+}} h(x)= \boxed{1}\) (b) \(\lim _{x \rightarrow 1^{-}} h(x)= \boxed{1}\) (c) \(\lim _{x \rightarrow 1} h(x)= \boxed{1}\)