1. **Problem statement:** (a) Find the domain of the function $$g(x) = \sqrt{1 - \sin(x)} \tan \left|\frac{3x}{x^2 + 1}\right|.$$ 2. **Domain analysis for $g(x)$:** - The square root requires the radicand to be non-negative: $$1 - \sin(x) \geq 0,$$ which means $$\sin(x) \leq 1.$$ Since the sine function never exceeds 1, this condition is true for all real $x$. - The tangent function demands we avoid points where $$\tan(\theta)$$ is undefined, i.e., where $$\theta = \frac{\pi}{2} + k\pi$$ for any integer $k$. Since the argument of tangent is the absolute value, which is always positive or zero, we find where $$\left|\frac{3x}{x^2 + 1}\right| = \frac{\pi}{2} + k\pi$$. 3. **Find $x$ where tangent is undefined:** Set $$\left|\frac{3x}{x^2 + 1}\right| = \frac{\pi}{2} + k\pi.$$ Because the denominator $x^2+1 > 0$ for all $x$, solve: $$\frac{3|x|}{x^2 + 1} = \frac{\pi}{2} + k\pi.$$ 4. **Check range of $\frac{3|x|}{x^2+1}$:** The function $$h(x) = \frac{3|x|}{x^2 + 1}$$ has maximum value at some positive $x$. Using calculus (critical points), find max: $$h'(x) = \frac{3(x^2+1) - 3|x|(2x)}{(x^2 + 1)^2} = 0,$$ which simplifies to find max at $$x=1$$ with $$h(1)=\frac{3(1)}{1+1} = \frac{3}{2} = 1.5.$$ 5. **Check if $1.5$ reaches values where tangent is undefined:** The smallest undefined value for tangent argument is $$\frac{\pi}{2} \approx 1.5708.$$ Since $1.5 < 1.5708$, it never reaches an undefined point. Therefore, no $x$ in real numbers makes tangent undefined. 6. **Conclusion on domain:** The domain is all real numbers $$\boxed{(-\infty, \infty)}$$. --- 7. **Problem (b)(i): Use the definition of a limit to show $$\lim_{x\to 0^+} \frac{1}{x} = \infty$$.** 8. **Definition approach:** We show for every large number $M > 0$, there exists $\delta > 0$ such that if $$0 < x < \delta,$$ then $$\frac{1}{x} > M.$$ 9. **Proof:** Given $M > 0$, choose $$\delta = \frac{1}{M}.$$ If $$0 < x < \delta,$$ then $$\frac{1}{x} > \frac{1}{\delta} = M.$$ Thus, as $x \to 0^+$, $\frac{1}{x} \to \infty.$ --- 10. **Problem (b)(ii):** Evaluate $$\lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^n.$$ 11. **Rewrite using exponentials:** Set $$a_n = \left(1 + \frac{2}{n}\right)^n,$$ then $$a_n = e^{n \ln \left(1 + \frac{2}{n}\right)}.$$ 12. **Using limit of logarithm:** As $$n \to \infty,$$ $$\ln \left(1 + \frac{2}{n}\right) \sim \frac{2}{n}$$ (since $$\ln(1+x) \approx x$$ for small $x$). Therefore, $$n \ln \left(1 + \frac{2}{n}\right) \to n \cdot \frac{2}{n} = 2.$$ 13. **Conclusion:** $$\lim_{n \to \infty} a_n = e^2.$$ --- 14. **Problem (c): Let $f : \mathbb{R} \to \mathbb{R}$ be defined by** $$f(x) = \begin{cases} x^2 - 2x + c, & x \leq 1, \\ dx - 1, & x > 1. \end{cases}$$ (Note: Since the problem snippet ends here without a question, no further answer is provided for part (c).)