QUESTION 6 Evaluate the following limits: (a) \( \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) \) (2) (b) \( \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} \) (c) Let \( f(x)=\frac{2 x+12}{|x+6|} \) (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 \( x=6 \). (3) (d) \( f(x)=\left\{\begin{array}{ll}\cos (x) & x<0 \\ 0 & x=0 \\ 1-x^{2} & x>0\end{array}\right. \) (i) Sketch the graph of \( f(x) \). (2) (ii) Is \( f(x) \) continuous at \( x=0 \) ? Give reason(s) for your answer. (3) [18]

Let's evaluate each part of Question 6 step by step. --- ### **6(a). \( \displaystyle \lim_{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) \)** **Solution:** To evaluate the limit of a product of functions as \( x \) approaches a specific value, we can compute the limits of each individual function and then multiply them, provided the limits exist. 1. **Compute \( \lim_{x \rightarrow 2} (x^2 + x) \):** \[ \lim_{x \rightarrow 2} (x^2 + x) = (2)^2 + 2 = 4 + 2 = 6 \] 2. **Compute \( \lim_{x \rightarrow 2} (3x^3 + 6) \):** \[ \lim_{x \rightarrow 2} (3x^3 + 6) = 3(2)^3 + 6 = 3(8) + 6 = 24 + 6 = 30 \] 3. **Multiply the two limits:** \[ \lim_{x \rightarrow 2} \left(x^{2}+x\right)\left(3x^{3}+6\right) = 6 \times 30 = 180 \] **Answer:** \( 180 \) --- ### **6(b). \( \displaystyle \lim_{u \rightarrow -2} \sqrt{u^{2} + 3u + 6} \)** **Solution:** To find this limit, substitute \( u = -2 \) directly into the function inside the square root, provided the expression under the square root remains non-negative. 1. **Plug \( u = -2 \) into the expression:** \[ u^2 + 3u + 6 = (-2)^2 + 3(-2) + 6 = 4 - 6 + 6 = 4 \] 2. **Take the square root:** \[ \sqrt{4} = 2 \] **Answer:** \( 2 \) --- ### **6(c). Let \( f(x) = \dfrac{2x + 12}{|x + 6|} \)** #### **6(c)(i). \( \displaystyle \lim_{x \rightarrow 6^{+}} f(x) \)** **Solution:** First, interpret \( x \rightarrow 6^{+} \) (approaching 6 from the right). 1. **Simplify \( f(x) \):** \[ f(x) = \frac{2x + 12}{|x + 6|} \] 2. **As \( x \rightarrow 6^{+} \), \( x + 6 \) is positive (since \( 6 + 6 = 12 > 0 \)), so \( |x + 6| = x + 6 \):** \[ f(x) = \frac{2x + 12}{x + 6} = \frac{2(x + 6)}{x + 6} = 2 \quad \text{(for } x \neq -6 \text{)} \] 3. **Thus, the limit is:** \[ \lim_{x \rightarrow 6^{+}} f(x) = 2 \] **Answer:** \( 2 \) #### **6(c)(ii). \( \displaystyle \lim_{x \rightarrow 6^{-}} f(x) \)** **Solution:** Now, interpret \( x \rightarrow 6^{-} \) (approaching 6 from the left). 1. **As \( x \rightarrow 6^{-} \), \( x + 6 \) is still positive (since \( 6 + 6 = 12 > 0 \)), so \( |x + 6| = x + 6 \):** \[ f(x) = \frac{2x + 12}{x + 6} = \frac{2(x + 6)}{x + 6} = 2 \quad \text{(for } x \neq -6 \text{)} \] 2. **Thus, the limit is:** \[ \lim_{x \rightarrow 6^{-}} f(x) = 2 \] **Answer:** \( 2 \) #### **6(c)(iii). Deduce \( \displaystyle \lim_{x \rightarrow 6} f(x) \)** **Solution:** Since both the right-hand limit and the left-hand limit as \( x \rightarrow 6 \) are equal and finite: \[ \lim_{x \rightarrow 6} f(x) = 2 \] **Answer:** \( 2 \) --- ### **6(d). \( f(x) = \begin{cases} \cos(x) & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 - x^{2} & \text{if } x > 0 \end{cases} \)** #### **6(d)(i). Sketch the graph of \( f(x) \)** **Solution:** To sketch the graph, consider each piece of the function separately. 1. **For \( x < 0 \):** \[ f(x) = \cos(x) \] - Plot the cosine curve for negative \( x \). 2. **At \( x = 0 \):** \[ f(0) = 0 \] - Place a point at \( (0, 0) \). Note that \( \cos(0) = 1 \) but \( f(0) = 0 \), indicating a point discontinuity. 3. **For \( x > 0 \):** \[ f(x) = 1 - x^{2} \] - This is a downward-opening parabola with vertex at \( (0, 1) \). 4. **Mark the behavior near \( x = 0 \):** - From the left, \( \cos(0) = 1 \), approaching \( 1 \) as \( x \) approaches 0 from the negative side. - From the right, as \( x \) approaches 0, \( 1 - x^{2} \) approaches 1. **Sketch Description:** - **Left of 0:** The standard cosine curve curving from \( ( -\pi, \cos(-\pi) = -1 ) \) upward to \( (0, 1) \). - **At 0:** A single point at \( (0, 0) \). - **Right of 0:** A parabola starting from just below \( (0, 1) \), descending downward as \( x \) increases. **[Since this is a text-based medium, here's a description. For an actual graph, refer to graphing tools or graph paper.]** **Answer:** *[As described above.]* #### **6(d)(ii). Is \( f(x) \) continuous at \( x = 0 \)? Give reason(s) for your answer.** **Solution:** To determine continuity at \( x = 0 \), three conditions must be satisfied: 1. **\( f(0) \) is defined:** - Yes, \( f(0) = 0 \). 2. **\( \lim_{x \rightarrow 0} f(x) \) exists:** - Compute left-hand and right-hand limits. **Left-hand limit (\( x \rightarrow 0^{-} \)):** \[ \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \cos(x) = \cos(0) = 1 \] **Right-hand limit (\( x \rightarrow 0^{+} \)):** \[ \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} (1 - x^{2}) = 1 - 0 = 1 \] - Since both one-sided limits are equal, \( \lim_{x \rightarrow 0} f(x) = 1 \). 3. **\( \lim_{x \rightarrow 0} f(x) = f(0) \):** - \( \lim_{x \rightarrow 0} f(x) = 1 \) vs. \( f(0) = 0 \). - They are not equal. **Conclusion:** Since \( \lim_{x \rightarrow 0} f(x) \neq f(0) \), the function \( f(x) \) is **not continuous** at \( x = 0 \). **Answer:** No, \( f(x) \) is not continuous at \( x = 0 \) because the limit as \( x \) approaches 0 is 1, which does not equal \( f(0) = 0 \). ---