Problem with limit and increasing function

Question

Let $f:(0,\infty) \to (0,\infty)$ be an increasing function such that $$\lim_{x \to \infty}f(x)=\infty \text { and } \lim_{x \to \infty} \frac {f(x+f(x))}{f(x)}=1.$$ Show that $$\lim_{x \to \infty} \frac {f(x)}{x}=0 \text { and } \lim_{x \to \infty} \frac {f(x+af(x))}{f(x)}=1,$$ for every $a \gt 0$.

Answer 1

First we will prove that $$ \lim_{x \to \infty} \frac{f(x)}x = 0. $$

Since $$ \lim_{x \to \infty} \frac {f(x+f(x))}{f(x)}=1, $$ we know that for any $\varepsilon > 0$ there is an $x_0$ such that $$ \tag{1} x \geq x_0 \implies \frac{f(x + f(x))}{f(x)} < 1 + \varepsilon. $$ We will use this to find an upper bound of $f(x)$ which holds for all $x \geq x_0$. Starting with $x_0$, define the sequence $\{x_n\}$ recursively by $$ x_{n+1} = x_n + f(x_n). $$ Note that $\{x_n\}$ is increasing, and that it is unbounded because $x_n \geq x_0 + nf(x_0)$. By setting $x = x_n$ in $(1)$, we can see that $$ f(x_{n+1}) < (1 + \varepsilon)f(x_n). $$

Taking this further, we find that

\begin{align} \frac{f(x_{n+1}) - f(x_n)}{x_{n+1} - x_n} &= \frac{f(x_{n+1}) - f(x_n)}{f(x_n)}\\ &< \frac{(1 + \varepsilon)f(x_n) - f(x_n)}{f(x_n)}\\ &= \varepsilon. \end{align} Therefore, when we plot $f$ at each $x_n$, the slope between two consecutive points is always less than $\varepsilon$, so $f$ grows no faster than a line of slope $\varepsilon$. More precisely, we have $$ f(x_n) \leq f(x_0) + \varepsilon(x_n - x_0). $$ Now, if $x \geq x_0$, then there is an $n$ such that $x_n \leq x < x_{n+1}$. So \begin{align} f(x) &\leq f(x_{n+1})\\ &< (1 + \varepsilon) f(x_n)\\ &\leq (1 + \varepsilon) (f(x_0) + \varepsilon(x_n - x_0))\\ &\leq (1 + \varepsilon) (f(x_0) + \varepsilon(x - x_0)). \end{align} Finally, we can use this upper bound to see that $$ \limsup_{x \to \infty} \frac{f(x)}x \leq \lim_{x \to \infty} \frac{(1 + \varepsilon) (f(x_0) + \varepsilon(x - x_0))}x = \varepsilon(1 + \varepsilon). $$ But since $\varepsilon$ can be chosen to be arbitrarily small, it must be that $$ \lim_{x \to \infty} \frac{f(x)}x = 0. $$

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Now we will prove that for all $a > 0$, $$ \lim_{x \to \infty} \frac{f(x + af(x))}{f(x)} = 1. \tag{2} $$

We will use induction. For the base case, note that if $a \leq 1$, then $$ 1 \leq \frac{f(x + af(x))}{f(x)} \leq \frac{f(x + f(x))}{f(x)} \to 1. $$ Now, let $x' = x + f(x)$ and assume that $(2)$ holds for $a$. Then, \begin{align} 1 \leq \frac{f(x + (a + 1)f(x))}{f(x)} &= \frac{f(x + (a + 1)f(x))}{f(x + f(x))} \cdot \frac{f(x + f(x))}{f(x)}\\ &= \frac{f(x' + af(x))}{f(x')} \cdot \frac{f(x + f(x))}{f(x)}\\ &\leq \frac{f(x' + af(x'))}{f(x')} \cdot \frac{f(x + f(x))}{f(x)} \to 1, \end{align}

So $(2)$ holds for $a + 1$. This concludes the induction step and proves that $(2)$ holds for all $a > 0$.

Answer 2

The limit $ \lim_{x\to \infty} \frac{f(x+f(x))}{f(x)}$ if of the form $\frac{\infty}{\infty}$. Thus, we can apply the LH Rule. Differentiating the denominator and numerator gives: $$ \lim_{x \to \infty} \frac{f'(x+f(x))}{f'(x)} (1+f'(x))$$ Which can be rearranged as: $$ \lim_{x \to \infty} \frac{f'(x+f(x))}{f'(x)}+f'(x+f(x))$$ We can now divide the problem into cases,

Case1: $f'(x)$ tends to some non-zero finite value (positive) $\lambda$

$\implies$ 1 = 1 + $\lambda$ $\implies$ $\lambda$ = 0

This case is rejected.

Case2: $f'(x)$ tends to infinite

In this case it is intuitive that the first term in the above sum is a positive indeterminate value and the second term is infinite(positive), their sum cannot be 1. Therefore, this case is rejected as well.

Having rejected the other possibilities, we can state that as x tends to infinite f'(x) tends to zero. (eg: a function such as ln(1+x))

The parts to show are now trivial (simple application of LH rule)