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: Read the first condition.
The condition $f(x)-x=\lambda$ for all positive $x$ tells us $f$ is just a shift, $f(x)=x+\lambda$.
Step 2: Plug this into the second condition.
Since $f(t)=t+\lambda$, we have $f(f(y))=f(y)+\lambda=(y+\lambda)+\lambda=y+2\lambda$, and $f(xy)+x=(xy+\lambda)+x$.
Step 3: Compare to pin down $\lambda$.
Setting $y+2\lambda=xy+\lambda+x$ and choosing $x=1$ gives $y+2\lambda=y+\lambda+1$, hence $\lambda=1$ and $f(x)=x+1$.
Step 4: Rewrite the required limit.
Then $f(x)=1+x$, so the limit is $\displaystyle\lim_{x\to0}\dfrac{(1+x)^{1/3}-1}{(1+x)^{1/2}-1}$.
Step 5: Use the small increment approximation.
For small $x$, $(1+x)^p-1\approx px$. So the numerator behaves like $\dfrac{x}{3}$ and the denominator like $\dfrac{x}{2}$.
Step 6: Take the ratio.
The limit equals $\dfrac{1/3}{1/2}=\dfrac{2}{3}$.
\[ \boxed{\dfrac{2}{3}} \]