# Grade 10 Optional Mathematics: Continuity
## Unit: Calculus | Chapter: Continuity and Discontinuity | Complete Solutions

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## 🔑 Key Concepts & Definitions

Before attempting the problems, it's important to understand the core concepts of limits and continuity.

### 1. Limits

* **Left-Hand Limit (LHL):** The value that a function `f(x)` approaches as `x` gets closer to a point `a` from the left side (values less than `a`). It is denoted as:
    $$ \lim_{x \to a^-} f(x) $$

* **Right-Hand Limit (RHL):** The value that a function `f(x)` approaches as `x` gets closer to a point `a` from the right side (values greater than `a`). It is denoted as:
    $$ \lim_{x \to a^+} f(x) $$

* **Existence of a Limit:** For the limit of a function `f(x)` to exist at `x = a`, the left-hand limit must equal the right-hand limit.
    $$ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) $$

### 2. Functional Value

* **Functional Value:** The value of the function `f(x)` at a specific point `x = a`. It is denoted as `f(a)`. This is the direct output of the function when `x` is exactly equal to `a`.

### 3. Continuity and Discontinuity

* **Continuity:** A function `f(x)` is said to be **continuous** at a point `x = a` if three conditions are met:
    1.  The left-hand limit exists: $\lim_{x \to a^-} f(x)$
    2.  The right-hand limit exists: $\lim_{x \to a^+} f(x)$
    3.  Both limits are equal to the functional value: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

    Essentially, the graph of the function has no breaks, jumps, or holes at that point.

* **Discontinuity:** A function is **discontinuous** at `x = a` if any of the three conditions for continuity are not met.

### 4. Intervals

* **Open Interval:** An interval that does not include its endpoints. It is denoted with parentheses, e.g., `(a, b)`.
* **Closed Interval:** An interval that includes its endpoints. It is denoted with square brackets, e.g., `[a, b]`.

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## 📚 Exercise Solutions

### Problem 1: Investigating Continuity
**Given the function f(x) = 3x + 2:**
**a) Find f(2.001), f(2.0001), f(1.999), and f(1.9999).**
**b) Find $\lim_{x \to 2^-} f(x)$, $\lim_{x \to 2^+} f(x)$, and f(2).**
**c) Is the function continuous at x = 2?**

**Solution:**
Here, $f(x) = 3x + 2$.

**a) Calculating values near x = 2:**
* $f(2.001) = 3(2.001) + 2 = 6.003 + 2 = 8.003$
* $f(2.0001) = 3(2.0001) + 2 = 6.0003 + 2 = 8.0003$
* $f(1.999) = 3(1.999) + 2 = 5.997 + 2 = 7.997$
* $f(1.9999) = 3(1.9999) + 2 = 5.9997 + 2 = 7.9997$
    
*As x approaches 2 from both sides, f(x) approaches 8.*

**b) Finding the limits and functional value:**

* **Left-Hand Limit (LHL):**
    $$ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (3x + 2) = 3(2) + 2 = 8 $$
* **Right-Hand Limit (RHL):**
    $$ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x + 2) = 3(2) + 2 = 8 $$
* **Functional Value:**
    $$ f(2) = 3(2) + 2 = 8 $$

**c) Checking for continuity:**
Since LHL = RHL = f(2) = 8, all three conditions for continuity are met.
**Yes, the function is continuous at x = 2.**

***

### Problem 2: Continuity of a Rational Function
**Given the function $f(x) = \frac{x^2 - 9}{x - 3}$:**
**a) Find f(2.999) and f(3.001).**
**b) Are f(2.999) and f(3.001) approximately equal?**
**c) Discuss the continuity of f(x) at x = 3.**

**Solution:**
Here, $f(x) = \frac{x^2 - 9}{x - 3}$.

**a) Calculating values near x = 3:**
* $f(2.999) = \frac{(2.999)^2 - 9}{2.999 - 3} = \frac{8.994001 - 9}{-0.001} = \frac{-0.005999}{-0.001} = 5.999$
* $f(3.001) = \frac{(3.001)^2 - 9}{3.001 - 3} = \frac{9.006001 - 9}{0.001} = \frac{0.006001}{0.001} = 6.001$

**b) Approximation:**
Yes, after approximation, both $f(2.999) \approx 6$ and $f(3.001) \approx 6$.

**c) Discussing continuity at x = 3:**
* **Limiting Value:**
    $$ \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \quad (\text{form } \frac{0}{0}) $$
    $$ = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 3+3=6 $$
* **Functional Value:**
    The function $f(3) = \frac{3^2 - 9}{3 - 3} = \frac{0}{0}$, which is undefined.

**Conclusion:** Since the functional value $f(3)$ is undefined, the function is **discontinuous** at $x = 3$.

***

### Problem 3: Examining Continuity of Piecewise Functions

**Examine the continuity or discontinuity of the following functions at the specified points.**

**a) $f(x) = \begin{cases} 2 - x^2, & \text{if } x \leq 2 \\ x - 4, & \text{if } x > 2 \end{cases}$ at $x = 2$**

**Solution:**
* **LHL:** $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2 - x^2) = 2 - 2^2 = -2$.
* **RHL:** $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x - 4) = 2 - 4 = -2$.
* **Functional Value:** $f(2) = 2 - 2^2 = -2$.

Since LHL = RHL = f(2) = -2, the function is **continuous** at $x = 2$.

**b) $f(x) = \begin{cases} x, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ x^2, & \text{if } x > 0 \end{cases}$ at $x = 0$**

**Solution:**
* **LHL:** $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x) = 0$.
* **RHL:** $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2) = 0^2 = 0$.
* **Functional Value:** $f(0) = 0$.

Since LHL = RHL = f(0) = 0, the function is **continuous** at $x = 0$.

**c) $f(x) = \begin{cases} \frac{2}{5 - x}, & \text{if } x \leq 3 \\ 5 - x, & \text{if } x > 3 \end{cases}$ at $x = 3$**

**Solution:**
* **LHL:** $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{2}{5 - x} = \frac{2}{5 - 3} = 1$.
* **RHL:** $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (5 - x) = 5 - 3 = 2$.

Since LHL $\neq$ RHL, the function is **discontinuous** at $x = 3$.

**d) $f(x) = \begin{cases} x^2 + 2, & \text{if } x < 5 \\ 6x - 3, & \text{if } x = 5 \\ 3x + 12, & \text{if } x > 5 \end{cases}$ at $x = 5$**

**Solution:**
* **LHL:** $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (x^2 + 2) = 5^2 + 2 = 27$.
* **RHL:** $\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x + 12) = 3(5) + 12 = 27$.
* **Functional Value:** $f(5) = 6(5) - 3 = 30 - 3 = 27$.

Since LHL = RHL = f(5) = 27, the function is **continuous** at $x = 5$.

***

### Problem 4: Finding an Unknown Value for Continuity
**A function is defined as $f(x) = \begin{cases} kx + 3, & \text{if } x \ge 2 \\ 3x - 1, & \text{if } x < 2 \end{cases}$. Find the value of k so that f(x) is continuous at x = 2.**

**Solution:**
For the function to be continuous at $x=2$, we must have LHL = RHL = f(2).

* **LHL:** (approaching 2 from the left, so $x < 2$)
    $$ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (3x - 1) = 3(2) - 1 = 5 $$
* **RHL:** (approaching 2 from the right, so $x > 2$)
    $$ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (kx + 3) = k(2) + 3 = 2k + 3 $$
* **Functional Value:** (at $x=2$)
    $$ f(2) = k(2) + 3 = 2k + 3 $$

Now, set LHL equal to RHL:
$$
\begin{aligned}
5 &= 2k + 3 \\
2 &= 2k \\
k &= 1
\end{aligned}
$$
**The value of k must be 1 for the function to be continuous at x = 2.**

***

### Problem 5: Finding an Unknown Value for Continuity (Rational Function)
**A function is defined as $f(x) = \begin{cases} \frac{2x^2 - 18}{x - 3}, & \text{for } x \neq 3 \\ k, & \text{for } x = 3 \end{cases}$. Find the value of k so that f(x) is continuous at x = 3.**

**Solution:**
For the function to be continuous at $x=3$, the limit of the function as $x \to 3$ must be equal to the functional value $f(3)$.

* **Limiting Value:**
    $$ \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{2x^2 - 18}{x - 3} $$
    This is in the indeterminate form $\frac{0}{0}$. We simplify the expression:
    $$ = \lim_{x \to 3} \frac{2(x^2 - 9)}{x - 3} = \lim_{x \to 3} \frac{2(x-3)(x+3)}{x-3} = \lim_{x \to 3} 2(x+3) = 2(3+3) = 12 $$
* **Functional Value:**
    $$ f(3) = k $$

To ensure continuity, we set the limiting value equal to the functional value:
$$ f(3) = \lim_{x \to 3} f(x) $$
$$ k = 12 $$
**The value of k must be 12 for the function to be continuous at x = 3.**

***

### Problem 6: Redefining a Function to be Continuous
**A function is defined as $f(x) = \begin{cases} 2x - 3, & \text{if } x < 2 \\ 2, & \text{if } x = 2 \\ 3x - 5, & \text{if } x > 2 \end{cases}$. Is f(x) continuous at x = 2? If not, how can it be made continuous?**

**Solution:**
Let's check the three conditions for continuity at $x=2$.

* **LHL:** $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x - 3) = 2(2) - 3 = 1$.
* **RHL:** $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 5) = 3(2) - 5 = 1$.
* **Functional Value:** $f(2) = 2$.

Here, LHL = RHL = 1, so the limit exists and is equal to 1. However, the functional value $f(2) = 2$.
Since $\lim_{x \to 2} f(x) \neq f(2)$, the function is **discontinuous** at $x=2$.

To make the function continuous, the functional value must be equal to the limit. We must redefine $f(2)$ to be 1.

The redefined continuous function is:
$$ f(x) = \begin{cases} 2x - 3, & \text{if } x < 2 \\ 1, & \text{if } x = 2 \\ 3x - 5, & \text{if } x > 2 \end{cases} $$