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

Step 1 ：First, we need to find the limit of the function $f(x)$ as $x$ approaches $-2$ from the left. This corresponds to the first piece of the piecewise function, which is $-x^{2}-3 x-1$. Substituting $x = -2$ into this equation, we get $1$.

Step 2 ：Next, we need to find the limit of the function $f(x)$ as $x$ approaches $-2$ from the right. This corresponds to the second piece of the piecewise function, which is $(x+1)^{2}$. Substituting $x = -2$ into this equation, we also get $1$.

Step 3 ：Finally, we need to find the limit of the function $f(x)$ as $x$ approaches $-2$ from both sides. If the limits from both sides are equal, then the limit exists and is that common value. If the limits from both sides are not equal, then the limit does not exist. In this case, since the limit from the left and the right are both $1$, the limit as $x$ approaches $-2$ is also $1$.

Step 4 ：So, the final answers are: (a) $\lim_{x \rightarrow -2^-} f(x) = \boxed{1}$, (b) $\lim_{x \rightarrow -2^+} f(x) = \boxed{1}$, and (c) $\lim_{x \rightarrow -2} f(x) = \boxed{1}$.