Question

Let $f\left(x\right)$ be a function defined on$(-a,a)$ with $a>0$. Amuse that $f\left(x\right)$ is continuous at $x=0$ and $\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(kx\right)}{x}=\alpha$, where $k\in (0,1)$ then compute $f^{\prime }\left(0^{+}\right)$ and $f^{\prime }\left(0^{-}\right)$, and comment upon the differentiablity of $f$ at $x=0$? Denote $\alpha$.

• A. $\underset{x\to o}{lim}\frac{f\left(x\right)-f\left(0\right)}{x}\cdot \frac{f\left(kx\right)-f\left(0\right)}{kx}\times k=\alpha$
• B. $\underset{x\to o}{lim}\frac{f\left(kx\right)-f\left(0\right)}{kx}\cdot \frac{f\left(kx\right)-f\left(0\right)}{kx}\times k=\alpha$
• C. $\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(kx\right)}{x}\cdot \frac{f\left(kx\right)-f\left(0\right)}{kx}\times k=\alpha$
• D. $\underset{x\to o}{lim}\frac{f\left(kx\right)-f\left(0\right)}{x}\cdot \frac{f\left(kx\right)-f\left(0\right)}{kx}\times k=\alpha$

Answer 1

The correct option is A $\underset{x\to o}{lim}\frac{f\left(x\right)-f\left(0\right)}{x}\cdot \frac{f\left(kx\right)-f\left(0\right)}{kx}\times k=\alpha$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}+f\left(x\right)$
$\underset{x\to 0^{-}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0}{lim}+f^{\prime }\left(x\right)$
$f^{\prime }\left(0^{=}\right)=f^{\prime }\left(0^{-}\right)$
$\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(kx\right)}{x}=\alpha$ ...(1)
$f^{\prime }\left(0^{+}\right)=\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x}$
$f^{\prime }\left(0^{-}\right)=\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{-x}$
From eqn (1)
$\underset{x\to 0}{lim}\frac{f\left(x\right)+f\left(0\right)-f\left(o\right)-f\left(kx\right)}{x}$
$\underset{x\to o}{lim}\frac{f\left(x\right)-f\left(0\right)}{x}\cdot \frac{f\left(kx\right)-f\left(0\right)}{kx}\times k=\alpha$