Mathematics - Continuity - Differentiability Question with Solution | TestHub

(i) Since $f(x)$ is continuous at $x=0$, ${\lim }_{x\to a}f(x)=f(0)$ - (1) Let $a\in \mathbb{R}$, then ${\lim }_{x\to a}f(x)={\lim }_{h\to 0}f(a+h)$

$$=\underset{h\to 0}{\lim }f(a)+f(h)$$

$$\begin{array}{rl}& =f(a)+\underset{h\to 0}{\lim }f(h)\\ & =f(a)+f(0)=f(a+0)\\ & =f(a)\end{array}$$

$\Rightarrow f(x)$ is continuous $\forall x\in \mathbb{R}$, as 'a' is arbitrary.

(ii) $\because f(x+y)=f(x)+f(y)\Rightarrow f(0)=0$. For any positive integer 'n', $f(n)=f(1+1+\dots +1(ntimes))=n\cdot f(1)$. For any negative integer $m$, we have $0=f(0)=f[m+(-m)]=f(m)+f(-m)$

$\Rightarrow f(m)=-f(-m)=-(-m)\cdot f(1)$

$$=m\cdot f(1)-(3)$$

(iii) Let $p/q$ be any rational number where 'q' is a positive integer and 'p' is any integer (positive, negative, or zero). Then $f\left(q\cdot \frac{p}{q}\right)=f\left(\frac{p}{q}+\frac{p}{q}+\dots +\frac{p}{q}(qtimes)\right)$

$$\begin{array}{rl}& =f\left(\frac{p}{q}\right)+f\left(\frac{p}{q}\right)+\dots +f\left(\frac{p}{q}\right)(qtimes)\\ & =q\cdot f\left(\frac{p}{q}\right)\\ \Rightarrow f(p)& =q\cdot f\left(\frac{p}{q}\right)----\left(4\right)\end{array}$$

But $f(p)=p\cdot f(1)$ from previous cases.

$\therefore p\cdot f(1)=q\cdot f(p/q)$

$\Rightarrow f(p/q)=\frac{p}{q}\cdot f(1)$ -----------(5)

(iv) Let 'x' be a real number. Since 'f' is continuous, $x_n\to x\Rightarrow f\left(x_n\right)\to f(x)$, where $⟨x_n⟩$ represents a sequence of rational numbers converging to 'x'. Since $x_n$ is a rational number, $f\left(x_n\right)=x_n\cdot f(1)$.

${\lim }_{n\to \infty }f\left(x_n\right)={\lim }_{n\to \infty }\left[x_n\cdot f(1)\right]$

$\Rightarrow {\lim }_{n\to \infty }f\left(x_n\right)=f(1)\left[{\lim }_{n\to \infty }{x}_n\right]$

$$=x\cdot f(1)-------\left(5\right)$$

$\therefore f(x)=x\cdot f(1)\forall x\in \mathbb{R}$. From all the above cases, we have $f(x)=kx,\forall x$, taking $f(1)=k$, where 'k' is a constant. (iii) and (iv) are obvious from $f(x)=kx$.