Mathematics - Limits Question with Solution | TestHub

Given relation is $f(x.f(y))=f(xy)+x$

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

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

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

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

So $f(x)=x+\lambda =x+1$ Hence ${\lim }_{x\to 0}\frac{(f(x){)}^{1/3}-1}{(f(x){)}^{1/2}-1}={\lim }_{x\to 0}\frac{(1+x{)}^{1/3}-1}{(1+x{)}^{1/2}-1}$

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