QUESTION 6 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 \( f(x)=\frac{2 x+12}{x+1 \mid} \) (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 \( x=6 \). (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 \( f(x) \). (ii) Is \( f(x) \) continuous at \( x=0 \) ? Give reason(s) for your answer. (3) [18]

Let's evaluate each part of the question step by step. --- ### **(a) Evaluate \( \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) \)** Since both \( x^2 + x \) and \( 3x^3 + 6 \) are **polynomial functions**, they are **continuous everywhere**. Therefore, we can directly substitute \( x = 2 \) into the expression. \[ \begin{align*} x^2 + x &= 2^2 + 2 = 4 + 2 = 6 \\ 3x^3 + 6 &= 3(2)^3 + 6 = 3(8) + 6 = 24 + 6 = 30 \\ \text{Limit} &= 6 \times 30 = 180 \end{align*} \] **Answer:** The limit is **180**. --- ### **(b) Evaluate \( \lim _{u \rightarrow -2} \sqrt{u^{2}+3u+6} \)** First, substitute \( u = -2 \) into the expression under the square root: \[ \begin{align*} u^2 + 3u + 6 &= (-2)^2 + 3(-2) + 6 = 4 - 6 + 6 = 4 \\ \sqrt{4} &= 2 \end{align*} \] **Answer:** The limit is **2**. --- ### **(c) Let \( f(x) = \frac{2x + 12}{|x + 1|} \)** #### **(i) Evaluate \( \lim _{x \rightarrow 6^{+}} f(x) \)** As \( x \) approaches 6 from the right (\( x \rightarrow 6^+ \)): \[ \begin{align*} 2x + 12 &= 2(6) + 12 = 12 + 12 = 24 \\ |x + 1| &= |6 + 1| = 7 \\ \text{Limit} &= \frac{24}{7} \end{align*} \] **Answer:** \( \lim _{x \rightarrow 6^{+}} f(x) = \frac{24}{7} \) #### **(ii) Evaluate \( \lim _{x \rightarrow 6^{-}} f(x) \)** As \( x \) approaches 6 from the left (\( x \rightarrow 6^- \)): \[ \begin{align*} 2x + 12 &= 2(6) + 12 = 24 \\ |x + 1| &= |6 + 1| = 7 \\ \text{Limit} &= \frac{24}{7} \end{align*} \] **Answer:** \( \lim _{x \rightarrow 6^{-}} f(x) = \frac{24}{7} \) #### **(iii) Deduce \( \lim f(x) \) at \( x = 6 \) using (i) and (ii)** Since both the left-hand limit and the right-hand limit as \( x \) approaches 6 are equal: \[ \lim_{x \to 6} f(x) = \frac{24}{7} \] **Answer:** \( \lim_{x \to 6} f(x) = \frac{24}{7} \) --- ### **(d) Let \( f(x) \) be defined as:** \[ f(x) = \begin{cases} \cos(x) & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 - x^2 & \text{if } x > 0 \end{cases} \] #### **(i) Sketch the graph of \( f(x) \)** **Description of the Graph:** 1. **For \( x < 0 \):** The graph follows \( \cos(x) \), which oscillates between -1 and 1. As \( x \) approaches 0 from the left, \( \cos(x) \) approaches 1. 2. **At \( x = 0 \):** The function has a single point at \( (0, 0) \). 3. **For \( x > 0 \):** The graph follows \( 1 - x^2 \), which is a downward-opening parabola starting at \( (0, 1) \). **Key Features:** - **Left of 0:** Continues as the cosine curve approaching \( \cos(0) = 1 \). - **At 0:** A distinct point at \( (0, 0) \), creating a **jump discontinuity**. - **Right of 0:** Begins at \( (0, 1) \) and decreases as \( x \) increases. **Graph Sketch:** While a visual sketch isn't possible here, imagine the cosine curve for negative \( x \), a point at the origin, and a downward parabola starting just above the origin. #### **(ii) Is \( f(x) \) continuous at \( x = 0 \)?** To determine continuity at \( x = 0 \), we check the following: 1. **Left-Hand Limit (\( x \to 0^- \)):** \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \cos(x) = \cos(0) = 1 \] 2. **Right-Hand Limit (\( x \to 0^+ \)):** \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1 - x^2) = 1 - 0 = 1 \] 3. **Function Value at 0:** \[ f(0) = 0 \] **Conclusion:** Since both the left-hand and right-hand limits as \( x \) approaches 0 are **1**, but \( f(0) = 0 \), the limits do not equal the function value. **Answer:** - **(i)** The graph consists of the cosine curve for \( x < 0 \), a point at \( (0, 0) \), and a downward parabola \( 1 - x^2 \) for \( x > 0 \). - **(ii)** \( f(x) \) is **not continuous** at \( x = 0 \) because the limit as \( x \) approaches 0 is 1, which does not equal \( f(0) = 0 \). --- **Summary of Answers:** - **(a)** 180 - **(b)** 2 - **(c)(i)** \( \frac{24}{7} \) - **(c)(ii)** \( \frac{24}{7} \) - **(c)(iii)** \( \frac{24}{7} \) - **(d)(i)** Graph as described above - **(d)(ii)** \( f(x) \) is not continuous at \( x = 0 \) since \( \lim_{x \to 0} f(x) = 1 \neq f(0) = 0 \)