Mathematics - Limits Question with Solution | TestHub

The given relation is $f(x\cdot f(y))=f(xy)+x$. Interchanging $x$ and $y$ in Eq. (1.56), we have $f(y\cdot f(x))=f(yx)+y$. Again, replacing $x$ with $f(x)$ in Eq. (1.56), we get $f(f(x)\cdot f(y))=f(y\cdot f(x))+f(x)$.

Therefore, Eqs. (1.56) - (1.58) imply $f(f(x)\cdot f(y))=f(xy)+y+f(x)$. Again, interchanging $x$ and $y$ in Eq. (1.59), we have $f(f(y)\cdot f(x))=f(yx)+x+f(y)$. Equations (1.59) and (1.60) imply $f(xy)+y+f(x)=f(yx)+x+f(y)$. Suppose $f(x)-x=f(y)-y=\lambda$. Substituting $f(x)=\lambda +x$ in Eq. (1.56), we have:

$$\begin{array}{r}x\cdot f(y)+\lambda =(xy+\lambda )+x\\ \Rightarrow x\cdot f(y)=xy+x\end{array}$$

Therefore, $x(y+\lambda )=xy+x[\because f(y)=\lambda +y]$.

$$\begin{array}{r}\Rightarrow \lambda x=x\\ \Rightarrow \lambda =1(\because x>0)\end{array}$$

So, $f(x)=x+\lambda =x+1$. Hence,

$$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{(f(x){)}^{1/3}-1}{(f(x){)}^{1/2}-1}& =\underset{x\to 0}{\lim }\frac{(1+x{)}^{1/3}-1}{(1+x{)}^{1/2}-1}\\ & =\underset{x\to 0}{\lim }\left(\frac{(1+x{)}^{1/3}-1}{1+x-1}\right)\cdot \left(\frac{1+x-1}{(1+x{)}^{1/2}-1}\right)\\ & =\frac{1/3}{1/2}=\frac{2}{3}\end{array}$$