Mathematics - Continuity - Differentiability Question with Solution | TestHub

The functional equation $f(x+y)=f(x)+f(y)$ for a continuous function $f:\mathbb{R}\to \mathbb{R}$ implies that $f(x)=cx$ for some constant $c$. Given $f(1)=2$, we have $c(1)=2$, so $c=2$. Thus, $f(x)=2x$. Now consider the function $g(x)={\lim }_{n\to \infty }\frac{(2x{)}^{2n}-1}{(2x{)}^{2n}+1}$. The behavior of this limit depends on the value of $|2x|$. 1. If$|2x|>1$, then $(2x{)}^{2n}\to \infty$ as $n\to \infty$, so $g(x)=\frac{1-0}{1+0}=1$. 2. If$|2x|<1$, then $(2x{)}^{2n}\to 0$ as $n\to \infty$, so $g(x)=\frac{0-1}{0+1}=-1$. 3. If$2x=1$, then $g(x)=\frac{1^{2n}-1}{1^{2n}+1}=\frac{1-1}{1+1}=0$. 4. If$2x=-1$, then $g(x)=\frac{(-1{)}^{2n}-1}{(-1{)}^{2n}+1}=\frac{1-1}{1+1}=0$. Thus,$g(x)$ changes its value at $2x=1$ (i.e., $x=1/2$) and at $2x=-1$ (i.e., $x=-1/2$). At $x=1/2$, $g(x)$ jumps from $-1$ to $0$ to $1$. At $x=-1/2$, $g(x)$ jumps from $1$ to $0$ to $-1$. Therefore, $g(x)$ is discontinuous at $x=1/2$ and $x=-1/2$. There are 2 points of discontinuity.