1. **Problem Statement:** (a) Find the domain of the function $$f(x) = \frac{x+3}{x^2 - 9}$$. (b) Evaluate the following limits: (i) $$\lim_{x \to 4} \frac{x-4}{\sqrt{x-2}}$$ (ii) $$\lim_{x \to +\infty} \sqrt{x^6 + 5x^3 - x^3}$$ (c) For the piecewise function $$f(x) = \begin{cases} \sqrt{-x}, & x < 0 \\ 3-x, & 0 \leq x < 3 \\ (x-3)^2, & x \geq 3 \end{cases}$$ (i) Find $$\lim_{x \to 0} f(x)$$ if it exists. (ii) Show that $$f(x)$$ is continuous at $$x=3$$. (iii) Sketch the graph of $$f(x)$$. 2. **Solution:** **(a) Domain of $$f(x) = \frac{x+3}{x^2 - 9}$$** - The denominator must not be zero: $$x^2 - 9 \neq 0$$ - Solve: $$x^2 \neq 9$$ $$x \neq \pm 3$$ - Thus, the domain is all real numbers except $$x = 3$$ and $$x = -3$$. **Domain:** $$\boxed{\{x \in \mathbb{R} : x \neq -3, x \neq 3\}}$$ **(b) Limits** (i) $$\lim_{x \to 4} \frac{x-4}{\sqrt{x-2}}$$ - Try direct substitution: $$\frac{4-4}{\sqrt{4-2}} = \frac{0}{\sqrt{2}} = 0$$ - Limit is directly 0. **Answer:** $$0$$ (ii) $$\lim_{x \to +\infty} \sqrt{x^6 + 5x^3 - x^3}$$ - Simplify inside the root: $$x^6 + 5x^3 - x^3 = x^6 + 4x^3$$ - Factor out $$x^6$$ inside the square root for large $$x$$: $$\sqrt{x^6(1 + \frac{4}{x^3})} = x^3 \sqrt{1 + \frac{4}{x^3}}$$ - As $$x \to +\infty$$, $$\frac{4}{x^3} \to 0$$, so: $$x^3 \times 1 = x^3$$, which tends to $$+\infty$$. **Answer:** $$+\infty$$ **(c) Piecewise function:** (i) $$\lim_{x \to 0} f(x)$$ - Left limit (x approaches 0 from below): $$\lim_{x \to 0^-} \sqrt{-x} = \sqrt{0} = 0$$ - Right limit (x approaches 0 from above): $$\lim_{x \to 0^+} 3 - x = 3 - 0 = 3$$ - Since left and right limits are not equal, the limit at 0 does not exist. **Answer:** Limit does not exist. (ii) Check continuity at $$x=3$$: - Left limit: $$\lim_{x \to 3^-} 3 - x = 3 - 3 = 0$$ - Right limit: $$\lim_{x \to 3^+} (x - 3)^2 = (3 - 3)^2 = 0$$ - Function value: $$f(3) = (3 - 3)^2 = 0$$ - Left limit, right limit, and function value are equal. **Conclusion:** $$f$$ is continuous at $$x=3$$. (iii) **Sketch:** - For $$x<0$$, $$f(x) = \sqrt{-x}$$ is the square root of positive values (since $$-x > 0$$), so a decreasing curve from $$x=0$$ to left. - For $$0 \leq x < 3$$, $$f(x) = 3 - x$$, a decreasing line from 3 at $$x=0$$ down to 0 at $$x=3$$. - For $$x \geq 3$$, $$f(x) = (x-3)^2$$, a parabola starting at 0 at $$x=3$$ and opening upwards. **Summary:** (a) Domain: $$x \neq \pm 3$$. (b)(i) Limit = 0. (b)(ii) Limit = +\infty. (c)(i) Limit at 0 does not exist. (c)(ii) Continuous at 3. (c)(iii) Piecewise graph with described behavior.