Consider the following function. \[ f(x)=\left\{\begin{array}{lll}x^{2}+x & \text { if } \quad x<-3 \\ 6 & \text { if } & x=-3 \\ 2 x-1 & \text { if } & x>-3\end{array}\right. \] Evaluate the given limits. If a limit does not exist, enter DNE. (a) \( \lim _{x \rightarrow-3^{-}} f(x)=\square \) (b) \( \lim _{x \rightarrow-3^{+}} f(x)=\square \) (c) \( \lim _{x \rightarrow-3} f(x)=\square \)

To evaluate the limits for the given piecewise function \( f(x) \), we will analyze the function as \( x \) approaches \(-3\) from the left and from the right. ### (a) \( \lim _{x \rightarrow-3^{-}} f(x) \) For \( x < -3 \), the function is defined as: \[ f(x) = x^2 + x \] We will calculate the limit as \( x \) approaches \(-3\) from the left: \[ \lim_{x \to -3^{-}} f(x) = \lim_{x \to -3^{-}} (x^2 + x) \] Substituting \( x = -3 \): \[ = (-3)^2 + (-3) = 9 - 3 = 6 \] Thus, \[ \lim _{x \rightarrow-3^{-}} f(x) = 6 \] ### (b) \( \lim _{x \rightarrow-3^{+}} f(x) \) For \( x > -3 \), the function is defined as: \[ f(x) = 2x - 1 \] We will calculate the limit as \( x \) approaches \(-3\) from the right: \[ \lim_{x \to -3^{+}} f(x) = \lim_{x \to -3^{+}} (2x - 1) \] Substituting \( x = -3 \): \[ = 2(-3) - 1 = -6 - 1 = -7 \] Thus, \[ \lim _{x \rightarrow-3^{+}} f(x) = -7 \] ### (c) \( \lim _{x \rightarrow-3} f(x) \) The limit as \( x \) approaches \(-3\) exists only if the left-hand limit and the right-hand limit are equal. We found: \[ \lim _{x \rightarrow-3^{-}} f(x) = 6 \] \[ \lim _{x \rightarrow-3^{+}} f(x) = -7 \] Since these two limits are not equal, the overall limit does not exist: \[ \lim _{x \rightarrow-3} f(x) = \text{DNE} \] ### Final Answers: (a) \( \lim _{x \rightarrow-3^{-}} f(x) = 6 \) (b) \( \lim _{x \rightarrow-3^{+}} f(x) = -7 \) (c) \( \lim _{x \rightarrow-3} f(x) = \text{DNE} \)