Question

Let \(f:\mathbb{R}^+\to\mathbb{R}^+\) be a function satisfying
\[ f(x)-x=\lambda \text{ constant},\quad \forall x\in\mathbb{R}^+ \] and
\[ f(f(y))=f(xy)+x,\quad \forall x,y\in\mathbb{R}^+. \] Then
\[ \lim_{x\to0}\frac{(f(x))^{1/3}-1}{(f(x))^{1/2}-1}= \]

• A. \(\dfrac{1}{3}\)
• B. \(0\)
• C. \(\dfrac{2}{3}\)
• D. \(1\)

Hint:

For limits involving \((1+x)^a-1\), use the standard approximation \((1+x)^a-1\sim ax\) as \(x\to0\).

Answer 1

The Correct Option is C

Step 1: Use the given condition \(f(x)-x=\lambda\).
Given,
\[ f(x)-x=\lambda \]
Therefore,
\[ f(x)=x+\lambda \]

Step 2: Apply the second functional equation.
Given,
\[ f(f(y))=f(xy)+x \]
Since \(f(t)=t+\lambda\), we get
\[ f(f(y))=f(y)+\lambda \]
Also,
\[ f(y)=y+\lambda \]
So,
\[ f(f(y))=y+2\lambda \]
Similarly,
\[ f(xy)+x=xy+\lambda+x \]
Thus,
\[ y+2\lambda=xy+\lambda+x \]

Step 3: Find \(\lambda\) using a suitable value.
Put \(x=1\). Then,
\[ y+2\lambda=y+\lambda+1 \]
\[ 2\lambda=\lambda+1 \]
\[ \lambda=1 \]
Therefore,
\[ f(x)=x+1 \]

Step 4: Substitute \(f(x)=x+1\) in the limit.
Now,
\[ \lim_{x\to0}\frac{(f(x))^{1/3}-1}{(f(x))^{1/2}-1} = \lim_{x\to0}\frac{(1+x)^{1/3}-1}{(1+x)^{1/2}-1} \]

Step 5: Use the standard limit formula.
We know that as \(x\to0\),
\[ (1+x)^a-1\sim ax \]
So,
\[ (1+x)^{1/3}-1\sim \frac{x}{3} \]
and
\[ (1+x)^{1/2}-1\sim \frac{x}{2} \]
Therefore,
\[ \lim_{x\to0}\frac{(1+x)^{1/3}-1}{(1+x)^{1/2}-1} = \frac{\frac{x}{3}}{\frac{x}{2}} \]
\[ =\frac{1}{3}\times2 \]
\[ =\frac{2}{3} \]

Step 6: Final conclusion.
Hence,
\[ \boxed{\frac{2}{3}} \]