QUESTION 6 Evaluate the following limits: (a) $$ \lim _{x ightarrow 2}\left(x^{2}+x ight)\left(3 x^{3}+6 ight) $$ (2) (b) $$ \lim _{u ightarrow-2} \sqrt{u^{2}+3 u+6} $$ (c) Let $$ f(x)=\frac{2 x+12}{|x+6|} $$ (i) $$ \lim _{x ightarrow 6^{+}} f(x) $$ (ii) $$ \lim _{x ightarrow 6^{-}} f(x) $$ (iii) Use (i) and (ii) to deduce $$ \lim f(x) $$ at $$ x=6 $$. (3) (d) $$ f(x)=\left\{\begin{array}{ll}\cos (x) & x0\end{array} ight. $$ (i) Sketch the graph of $$ f(x) $$. (2) (ii) Is $$ f(x) $$ continuous at $$ x=0 $$ ? Give reason(s) for your answer. (3) [18]

Question

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

UpStudy AI · Accepted Answer

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 $$ f(0) = 0 $$. --- **6(a)**: The limit is 180. **6(b)**: The limit is 2. **6(c)(i)**: The limit as $$ x $$ approaches 6 from the right is 2. **6(c)(ii)**: The limit as $$ x $$ approaches 6 from the left is 2. **6(c)(iii)**: The overall limit as $$ x $$ approaches 6 is 2. **6(d)(i)**: The graph of $$ f(x) $$ consists of the cosine curve for $$ x < 0 $$, a single point at $$ (0, 0) $$, and the parabola $$ 1 - x^{2} $$ for $$ x > 0 $$. **6(d)(ii)**: The function $$ f(x) $$ is not continuous at $$ x = 0 $$ because the limit as $$ x $$ approaches 0 is 1, but $$ f(0) = 0 $$.

QUESTION 6

Evaluate the following limits:
(a)
(2)
(b)
© Let
(i)
(ii)
(iii) Use (i) and (ii) to deduce at .
(3)
(d)
(i) Sketch the graph of .
(2)
(ii) Is continuous at ? Give reason(s) for your answer.
(3)
[18]

Solución de inteligencia artificial de Upstudy

Responder

Solución

6(a).

6(b).

6©. Let

6©(i).

6©(ii).

6©(iii). Deduce

**6(d). ( f(x) = \begin{cases}

6(d)(i). Sketch the graph of

6(d)(ii). Is continuous at ? Give reason(s) for your answer.

Beyond the Answer

preguntas relacionadas

Latest Calculus Questions

QUESTION 6 Evaluate the following limits: (a) (2) (b) © Let (i) (ii) (iii) Use (i) and (ii) to deduce at . (3) (d) (i) Sketch the graph of . (2) (ii) Is continuous at ? Give reason(s) for your answer. (3) [18]