QUESTION 6 6.1 Evaluate the following limits: (a) \( \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) \) (b) \( \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} \) (c) Let \( f(x)=\frac{2 x+12}{|x+6|} \) Determine: (i) \( \lim _{x \rightarrow 6^{+}} f(x) \) (ii) \( \lim _{x \rightarrow 6^{-}} f(x) \) (iii) Use (i) and (ii) to deduce \( \lim f(x) \) at \( x=6 \). 6.2 Let \[ f(x)=\left\{\begin{array}{cll} \cos x & \text { if } & x<0 \\ 0 & \text { if } & x=0 \\ 1-x^{2} & \text { if } & x>0 \end{array}\right. \]

Sure, let's evaluate each part of **Question 6.1** step by step. --- ### **6.1 Evaluate the following limits:** #### **(a) \( \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) \)** **Solution:** The function \( \left(x^{2}+x\right)\left(3 x^{3}+6\right) \) is a polynomial, which is continuous everywhere. Therefore, we can directly substitute \( x = 2 \) into the function to find the limit. 1. **Calculate \( x^2 + x \) at \( x = 2 \):** \[ 2^2 + 2 = 4 + 2 = 6 \] 2. **Calculate \( 3x^3 + 6 \) at \( x = 2 \):** \[ 3(2)^3 + 6 = 3(8) + 6 = 24 + 6 = 30 \] 3. **Multiply the two results:** \[ 6 \times 30 = 180 \] **Answer:** \[ \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) = 180 \] --- #### **(b) \( \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} \)** **Solution:** First, evaluate the expression inside the square root at \( u = -2 \): 1. **Calculate \( u^2 + 3u + 6 \) at \( u = -2 \):** \[ (-2)^2 + 3(-2) + 6 = 4 - 6 + 6 = 4 \] 2. **Take the square root of the result:** \[ \sqrt{4} = 2 \] Since the function inside the limit is continuous at \( u = -2 \), the limit is simply the value of the function at that point. **Answer:** \[ \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} = 2 \] --- #### **(c) Let \( f(x)=\frac{2 x+12}{|x+6|} \). Determine:** ##### **(i) \( \lim _{x \rightarrow 6^{+}} f(x) \)** **Solution:** When \( x \) approaches \( 6 \) from the right (\( x \rightarrow 6^{+} \)): 1. **Evaluate the denominator \( |x + 6| \):** \[ x + 6 \rightarrow 6 + 6 = 12 \quad (\text{since } x > 6) \] \[ |x + 6| = 12 \] 2. **Evaluate the numerator \( 2x + 12 \) at \( x = 6 \):** \[ 2(6) + 12 = 12 + 12 = 24 \] 3. **Compute the limit:** \[ \frac{24}{12} = 2 \] **Answer:** \[ \lim _{x \rightarrow 6^{+}} f(x) = 2 \] --- ##### **(ii) \( \lim _{x \rightarrow 6^{-}} f(x) \)** **Solution:** When \( x \) approaches \( 6 \) from the left (\( x \rightarrow 6^{-} \)): 1. **Evaluate the denominator \( |x + 6| \):** \[ x + 6 \rightarrow 6 + 6 = 12 \quad (\text{since } x < 6, \text{ but } x + 6 > 0) \] \[ |x + 6| = 12 \] 2. **Evaluate the numerator \( 2x + 12 \) at \( x = 6 \):** \[ 2(6) + 12 = 12 + 12 = 24 \] 3. **Compute the limit:** \[ \frac{24}{12} = 2 \] **Answer:** \[ \lim _{x \rightarrow 6^{-}} f(x) = 2 \] --- ##### **(iii) Use (i) and (ii) to deduce \( \lim_{x \rightarrow 6} f(x) \) at \( x=6 \).** **Solution:** Since both the right-hand limit (\( 2 \)) and the left-hand limit (\( 2 \)) are equal, the two-sided limit exists and is equal to this common value. **Answer:** \[ \lim_{x \rightarrow 6} f(x) = 2 \] --- ### **Summary of Answers for Question 6.1:** - **(a)** The limit is **180**. - **(b)** The limit is **2**. - **(c)** - **(i)** \( \lim _{x \rightarrow 6^{+}} f(x) = 2 \) - **(ii)** \( \lim _{x \rightarrow 6^{-}} f(x) = 2 \) - **(iii)** \( \lim_{x \rightarrow 6} f(x) = 2 \) --- If you have any further questions or need assistance with **Question 6.2**, feel free to ask!