 Solution

To prove boundedness on $[a,b]$, suppose instead that $f$ is unbounded and choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence converging to $x_{\ast }\in [a,b]$. Continuity then gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j$. Hence $f$ is bounded.

Let $M={\sup }_{x\in [a,b]}f(x)$ and $m={\inf }_{x\in [a,b]}f(x)$. Choose $d_n$ with $M-1/n<f(d_n)\le M$. Another application of Bolzano-Weierstrass theorem yields a subsequence tending to $d\in [a,b]$, and continuity gives $f(d)=M$. Applying the same argument to $-f$ gives $c\in [a,b]$ with $f(c)=m$. ThusThis proves the extreme value theorem here, including both attainment assertions.

For $g$, its two limits supply $R>0$ such that $|g(x)|<1$ whenever $|x|>R$. On $[-R,R]$, continuity and the result just proved give a finite bound $K$. Consequently $|g(x)|\le \max \{K,1\}$ on the whole real line.

For the final value assertion, if $c=g(a)$, take $x=a$. Otherwise $0<c<g(a)$. The positive-infinity limit gives a point $b>a$ with $|g(b)|<c$, so $g(b)<c<g(a)$. The intermediate value theorem on $[a,b]$ now gives $x\in (a,b)$ withPositivity of $c$ is essential to this tail argument; a function can tend to zero without ever attaining zero.