I am solving real analysis notes of my friend where i am having doubt in the below problem.
Show that if $x>0$, $\log(1+x)>\frac x{1+x}$ and hence prove that $x^{-1}\log(1+x)$ decreases monotonically as $x$ increases from $0$ to $\infty$.
Solution: Let $f(x)=\log(1+x)-\frac x{1+x}$, then $$f'(x)=\frac1{1+x}-\frac{(1+x)\cdot1-x}{(1+x)^2}=\frac x{(1+x)^2}>0\ (\because x>0)$$ $$\therefore x>0\implies f'(x)>0$$ i.e $f$ is an increasing when $x>0$ and $f(x)>f(0)$.
In the text, the author has directly stated that $f(x) > f(0)$ even though $f'(x) > 0 $ only when $x>0$. My understanding of the problem is as follows.
Technically speaking , given that $f'(x) > 0$ when $x > 0$ , $\implies$ $f'(x) > f(0^+)$ . Since the function is continuous at $x = 0$ , hence, $\;f(0^+) = f(0) $ which leads to $f(x) > f(0)$.
My questions to the community are as follows :-
- Does my reasoning holds good ?
- If someone can explain / reason what is happening to the function $f(x)$ at $x=0$.
Your reasoning is not correct. Argue as follows: Suppose $x<y$. The only case when we have difficulty in showing that $f(x) <f(y)$ is when $x=0$. In this case prove by contradiction. Suppose $f(0)(=f(x))\geq f(y)$. We have $f(\frac 1 n) < f(\frac y 2) $ for all $n$ such that $\frac 1 n <\frac y 2$. Letting $n \to \infty$ and using continuity at $0$ we get $f(0) \leq f(\frac y 2) <f(y)$. [Note that $<$ becomes $\leq $ in the limit. This is important]. We have now arrived at a contradiction.