Evhate the sollowine limits (a) \( \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{2}+6\right) \) (b) \( \lim _{x \rightarrow-2} \sqrt{x^{2}+3 x+6} \)

Certainly! Let's evaluate each of the given limits step by step. --- ### Problem (a) \[ \lim_{x \rightarrow 2} \left(x^{2} + x\right)\left(3x^{2} + 6\right) \] **Step 1: Direct Substitution** Since the functions involved are polynomials, they are continuous everywhere. This means we can directly substitute \( x = 2 \) into the expression. **Step 2: Substitute \( x = 2 \)** \[ \begin{align*} \left(2^{2} + 2\right)\left(3 \times 2^{2} + 6\right) &= \left(4 + 2\right)\left(3 \times 4 + 6\right) \\ &= (6)(12 + 6) \\ &= 6 \times 18 \\ &= 108 \end{align*} \] **Answer:** \[ \lim_{x \rightarrow 2} \left(x^{2} + x\right)\left(3x^{2} + 6\right) = 108 \] --- ### Problem (b) \[ \lim_{x \rightarrow -2} \sqrt{x^{2} + 3x + 6} \] **Step 1: Direct Substitution** First, check if direct substitution is possible by plugging \( x = -2 \) into the expression inside the square root. \[ x^{2} + 3x + 6 = (-2)^{2} + 3(-2) + 6 = 4 - 6 + 6 = 4 \] Since the expression inside the square root is 4, which is non-negative, the square root is defined. **Step 2: Calculate the Square Root** \[ \sqrt{4} = 2 \] **Answer:** \[ \lim_{x \rightarrow -2} \sqrt{x^{2} + 3x + 6} = 2 \] --- **Summary:** - **Problem (a)** evaluates to **108**. - **Problem (b)** evaluates to **2**. If you have any further questions or need additional explanations, feel free to ask!