1. Take $c \not \in \mathbb{Q}$. Choose $x_{n} \in \mathbb{Q}$ such that $x_{n} \to c$. Since $f$ is continuous, $f(x_{n}) \to f(c) \implies g(x_{n}) \to f(c)$ since $f$ and $g$ are equal at rational points. But $g(x_{n}) \to g(c)$ so $f(c) = g(c)$ by uniqueness of limits. 7. Let $f:(-\infty,0] \to \mathbb{R}$ and $g:[0,\infty) \to \mathbb{R}$ both be continuous on their entire domain. Define the function $h: \mathbb{R} \to \mathbb{R}$ by $h(x)=\begin{cases} f(x) & x \leq 0 \\ g(x) & x > 0 \end{cases}$Firstly suppose $h$ is continuous at $x=0$. Now choose $x_{n}>0$ such that $x_{n} \to 0$, Then $g(x_{n}) = h(x_{n}) \to h(0) = f(0)$ since $h$ is continuous. Also $g(x_{n}) \to g(0)$ as $n \to \infty$ since $g$ is continuous in its domain. So $f(0) = g(0)$ by uniqueness of limits. Conversely, suppose $f(0) \neq g(0)$. Choose $x_{n} > 0$ such that $x_{n} \to 0$, Then $h(x_{n}) = g(x_{n}) \to g(0) \neq f(0) = h(0)$ since $g$ is continuous in its domain. Hence $h$ is discontinuous at $x=0$. 14. WLOG suppose $L>0$. Take $\phi \in (0,L)$ and $\epsilon=L-\phi >0$. Choose $a>R$ then $f(a)-L > -\epsilon = -L + \phi \implies \phi<f(a)$ Since $f$ is continuous on $[0,a]$ (also $f(a)>\phi>0=f(0)$) and $f(0)=0<\phi<f(a)$, by IVT, $\exists x \in (0,a) \subset(0,\infty)$ such that $f(x) = \phi$. Similarly for $L<0$ choose $\epsilon=\phi-L$. 19. Choose $\epsilon>0$ then $\exists R>0$ such that $|f(x)-L| < \epsilon$ for all $x>R$. So, using triangle inequality $|f(x)| \leq |f(x)-L|+|L| < \epsilon+ |L|$ for $x \in (R,\infty)$. Applying EVT to $f:[0,R]\to \mathbb{R}$ we have that $\exists m>0$ such that $|f(x)|<m$ for all $x\in [0,R].$ So $|f(x)|<\max(\epsilon+|L|, m)$ for all $x\in[0,\infty]$ which shows that $f$ is bounded. If $m= \max(\epsilon+|L|,m)$ then by EVT, $f$ attains its bounds. However, if $\epsilon+|L|= \max(\epsilon+|L|, m)$ then it does necessarily attain its bounds. For example take $f=\frac{1}{x+1 } \neq 0$, its lower bound.