Main Question [5 marks]

Show that $f(x) = 1 - \sqrt{1 - x^2}$ is continuous on $[-1,1]$

A function is continuous at a point $c$ if $\lim_{x \to c} f(x) = f(c)$. A function is continuous on an interval if it is continuous at every point in that interval.

Proof:

Let $c \in [-1, 1]$. We need to show that $\lim_{x \to c} f(x) = f(c)$.

Step a: The function $f(x) = 1 - \sqrt{1-x^2}$

• $g(x) = x^2$ is a polynomial, hence continuous everywhere.
• $h(x) = 1 - x^2$ is a polynomial, hence continuous everywhere.
• For $x \in [-1,1]$, we have $1 - x^2 \geq 0$, so $\sqrt{1-x^2}$ is defined.
• $\sqrt{x}$ is continuous on its domain $[0, \infty)$.

Step b: By the composition rule, $\sqrt{1-x^2} = \sqrt{h(x)}$ is continuous on $[-1,1]$ since $h(x) \geq 0$ on this interval.

Step c: Since the difference of continuous functions is continuous:

$$f(x) = 1 - \sqrt{1-x^2}$$

is continuous on $[-1,1]$ (constant function $1$ minus a continuous function $\sqrt{1-x^2}$).

Step d: At the endpoints:

• At $x = -1$: $\lim_{x \to -1^+} f(x) = 1 - \sqrt{1-1} = 1 = f(-1)$ ✓
• At $x = 1$: $\lim_{x \to 1^-} f(x) = 1 - \sqrt{1-1} = 1 = f(1)$ ✓

Conclusion: Since $f(x)$ is continuous at every interior point and at both endpoints (one-sided continuity), $f(x)$ is continuous on $[-1,1]$.

---

Sub-question 1 [5 marks]

Evaluate: $\lim_{x \to \infty} \sqrt{x}(\sqrt{x-2} - \sqrt{x})$

Strategy: Rationalize the expression by multiplying and dividing by the conjugate.

Step a: Multiply and divide by the conjugate:

$$\sqrt{x}(\sqrt{x-2} - \sqrt{x}) \times \frac{\sqrt{x-2} + \sqrt{x}}{\sqrt{x-2} + \sqrt{x}}$$

Step b: Simplify the numerator using $(a-b)(a+b) = a^2 - b^2$:

$$= \sqrt{x} \cdot \frac{(x-2) - x}{\sqrt{x-2} + \sqrt{x}}$$

$$= \sqrt{x} \cdot \frac{-2}{\sqrt{x-2} + \sqrt{x}}$$

Step c: Simplify:

$$= \frac{-2\sqrt{x}}{\sqrt{x-2} + \sqrt{x}}$$

Step d: Divide numerator and denominator by $\sqrt{x}$:

$$= \frac{-2}{\frac{\sqrt{x-2}}{\sqrt{x}} + 1} = \frac{-2}{\sqrt{\frac{x-2}{x}} + 1}$$

$$= \frac{-2}{\sqrt{1 - \frac{2}{x}} + 1}$$

Step e: Apply the limit as $x \to \infty$:

As $x \to \infty$, $\frac{2}{x} \to 0$, so $\sqrt{1 - \frac{2}{x}} \to \sqrt{1} = 1$

$$= \frac{-2}{1 + 1} = \frac{-2}{2}$$

$$\boxed{\lim_{x \to \infty} \sqrt{x}(\sqrt{x-2} - \sqrt{x}) = -1}$$