Question

If $f\left(x\right)=x\cdot \left(\frac{a^{1/x}-a^{-1/x}}{a^{1/x}+a^{-1/x}}\right),x\ne 0(a>0,)f\left(0\right)=0$ then

• A. f is differentiable at x = 0
• B. f is not differentiable at x = 0
• C. f is not continuous at x = 0
• D. None of these

Answer 1

Answer: (B)

f is not differentiable at x = 0

Given $f\left(x\right)=x\left(\frac{a^{1/x}-a^{-1/x}}{a^{1/x}+a^{-1/x}}\right)$
Let's check whether $f$ is differentiable at $x=0$
$RHD=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$\Rightarrow RHD=\underset{h\to 0}{lim}\frac{h\times \left(\frac{a^{1/h}-a^{-1/h}}{a^{1/h}+a^{-1/h}}\right)-0}{h}$
$RHD=\underset{h\to 0}{lim}\frac{a^{2/h}-1}{a^{2/h}+1}$
$RHD=\underset{h\to 0}{lim}\frac{a^{2/h}(1-a^{-2/h})}{a^{2/h}(1+a^{-2/h})}$
$\Rightarrow RHD=1$
$LHD=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$\Rightarrow LHD=\underset{h\to 0}{lim}\frac{-h\times \left(\frac{a^{-1/h}-a^{1/h}}{a^{-1/h}+a^{1/h}}\right)-0}{-h}$
$LHD=\underset{h\to 0}{lim}\frac{1-a^{2/h}}{1+a^{2/h}}$
$LHD=\underset{h\to 0}{lim}\frac{a^{2/h}(a^{-2/h}-1)}{a^{2/h}(a^{-2/h}+1)}$
$\Rightarrow LHD=-1$
Hence, $LHD\ne RHD$ at $x=0$
Hence, f(x) is not differentiable at $x=0$
Now, let's check continuity at $x=0$
$RHL=\underset{h\to 0^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}h\left(\frac{a^{1/h}-a^{-1/h}}{a^{1/h}+a^{-1/h}}\right)$
$=\underset{h\to 0}{lim}h\left(\frac{a^{2/h}-1}{a^{2/h}+1}\right)$
$=\underset{h\to 0}{lim}h\left(\frac{a^{2/h}(1-a^{-2/h})}{a^{2/h}(1+a^{-2/h})}\right)$
$RHL=0$
$LHL=\underset{h\to 0^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}(-h)\left(\frac{a^{-1/h}-a^{1/h}}{a^{-1/h}+a^{1/h}}\right)$
$=\underset{h\to 0}{lim}(-h)\left(\frac{(1-a^{2/h})}{(1+a^{2/h})}\right)$
$=\underset{h\to 0}{lim}(-h)\left(\frac{a^{2/h}(a^{-2/h}-1)}{a^{2/h}(a^{-2/h}+1)}\right)$
$\Rightarrow LHL=0$
Hence, $LHL=RHL=f\left(0\right)$
Hence, $f\left(x\right)$ is continuous at $x=0$