Prove that $\lim_{x \rightarrow \infty} \frac{f(x)}{x}= \alpha$ when $f$ is continuous and $\lim_{x \rightarrow \infty} (f(x+1) - f(x)) = \alpha$

Question

Prove that $\lim_{x \rightarrow \infty} \frac{f(x)}{x}= \alpha$ when $f$ is continuous and $\lim_{x \rightarrow \infty} (f(x+1) - f(x)) = \alpha$

The given hint suggests to consider the case of $\alpha=0$.

Given $\varepsilon \gt 0$, there exists $M$ such that $x \ge M $ implies that $|f(x+1) - f(x)| \lt \varepsilon$. Then for any $y$ which is larger than $M$, we can write it as $y=x+N$ where some $x \in [M, M+1]$.

And $f$ is continuous on $[M, M+1]$, it has a maximum $M_0$ on this interval, i.e., $|f(x)| \le M_0$ on this interval. Then $|f(x+N)| \le M_0 + N\varepsilon$. So $|\frac{f(x+N)}{x+N}| \le \frac{M_0 + N\varepsilon}{x+N}$. Then as $N \rightarrow \infty$, $\lim_{y \rightarrow \infty} \frac{f(y)}{y} \le \varepsilon$, which means that $\lim_{y \rightarrow \infty} \frac{f(y)}{y} =0$.

Is this argument valid? And I’d like to know whether there is a simple argument.

Answer 1

I am thinking in terms of your argument. I think what might be confusing in your answer is the simultaneous variation of $x \in [M,M+1]$ and $N \in \mathbb{N}$ as $y \to \infty $

Let $y \gt M+1$ . Then $\exists x_y\in [M,M+1], N_y\in \mathbb{N}$ such that

$y=x_y+N_y$ Then

$|f(y)-f(x_y)|=|f(x_y+N_y)-f(x_y)|$

$=|f(x_y+N_y)-f(x_y+N_y-1)+f(x_y+N_y-1)-f(x_y+N_y-2)...+f(x_y+1)-f(x_y)|\le N_y \epsilon$ (By Triangle Inequality)

So $ |f(y)|-|f(x_y)| \le |f(y)-f(x_y)|\le N_y \epsilon$

(since $|a-b| \ge ||a|-|b|| \ge |a|-|b|$)

which gives

$|f(y)| \le M_0+N_y \epsilon$

Now, $\big| \frac {f(y)}{y}\big| \le \frac {M_0/N_y+\epsilon}{x_y/N_y+1}$

Since $x_y$ is bounded , we must have as $N_y \to \infty $ , $\displaystyle\lim_{y\to \infty} \big| \frac {f(y)}{y}\big| \le \epsilon$

Answer 2

Apologies if this is an overly tortured derivation. Let me know if there are any hand-wavey parts.

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We know that for any $\epsilon > 0$ there exists $x_{\epsilon}$ such that $\left| f(x+1) - f(x)-\alpha\right| < \epsilon\;\forall x>x_{\epsilon}$.

For brevity, take $x'=\max\{x_{\epsilon},1\}$

This implies $ -\epsilon <f(x'+\delta+1) - f(x'+\delta)-\alpha < \epsilon\;\;\forall \delta > 0$

Dividing by $x'+\delta$ we get:

$$\frac{-\epsilon}{x'+\delta} < \frac{f(x'+\delta+1) - f(x'+\delta)-\alpha}{x'+\delta} < \frac{\epsilon}{x'+\delta}$$

Taking the limit as $\delta \to \infty$ we get:

$$0 < \frac{f(x'+\delta+1) - f(x'+\delta)-\alpha}{x'+\delta} < 0$$

Therefore,

$$\lim_{x\to \infty} \frac{f(x+1) - f(x)-\alpha}{x} = 0$$

This implies that $f(x) = c x + o(x)$ for some $c$

If we assume $c\neq 0$, then $c>\alpha \implies \lim_{x\to \infty} \frac{f(x+1) - f(x)-\alpha}{x} > 0$ and analogously for $c<\alpha$, therefore, $c=\alpha$.

If we assume $c=0$ then $f(x) = o(x)$ and $\lim_{x\to\infty} f(x+1) - f(x) = 0$. If $\alpha =0$ this works out fine, but if not then we have a contradiction with a prior assumption.

Therefore, $c=\alpha$ and $f(x) = \alpha x + o(x) \implies \lim_{x\to\infty}\frac{f(x)}{x} = \alpha$