# Show that $\lim_{x \rightarrow \infty} f'(x)<1$ implies $f(x_0)<x_0$ for some $x_0$

Let $$f:[0,\infty)\rightarrow R$$ be a continuously differentiable function. Show that if

$$\lim_{x \rightarrow \infty} f'(x)<1$$

then

$$f(x_0) for some $$x_0$$ large enough. (An example of a function that satisfies these assumption is $$f(x) = \sqrt x$$). I am struggling with the proof. I've tried with the mean value theorem:

Choose $$0. Then by the MVT there esists a number $$c \in (x,y)$$ such that

$$\dfrac{f(y)-f(x)}{y-x} = f'(c)$$

But at this point I got stuck... The other info I have is that there exists a number $$M>0$$ such that if $$x>M$$ then $$f'(x)<1$$ (the limit condition stated in the hp). Is there a way to combine these two facts in order to prove what I want?

• I added the "real-analysis" tag to your post. Cheers! – Robert Lewis Oct 4 '18 at 16:48

## 3 Answers

Take $$r\in\left(\lim_{x\to\infty}f'(x),1\right)$$. Since $$\lim_{x\to\infty}f'(x), there is a $$M>0$$ such that $$x\geqslant M\implies f'(x). Then, when $$x>M$$, by the mean value theorem$$\frac{f(x)-f(M)}{x-M}=f'(c)for some $$c$$ between $$M$$ and $$x$$. So,$$x>M\implies f(x)But, if $$x$$ is large enough, we have $$f(M)+r(x-M), since $$0, and therefore, if $$x$$ is large enough, we have $$f(x).

• The first two sentences of your answer make no sense. $\lim f > r$ no given in the problem, also how you find the $M$ ? – Mikey Spivak Oct 4 '18 at 16:51
• @Neymar THere was a typo in the second sentence: I wrote $f$ intead of $f'$. What's the problem with the first sentence? Since $\lim_{x\to\infty}f'(x)<1$, I can consider the interval $\left(\lim_{x\to\infty}f'(x),1\right)$. – José Carlos Santos Oct 4 '18 at 16:57
• i dont understand really its too terse the way you write – Mikey Spivak Oct 4 '18 at 17:43
• Thanks but I don't understand the last statement: "if x is large enough, we have $f(M)+r(x−M)<x$ since $0<r<1$" – Alessandro Oct 4 '18 at 17:50
• Note that$$f(M)+r(x-M)<x\iff f(M)-rM<(1-r)x,$$which is true if $x$ is large enough. – José Carlos Santos Oct 4 '18 at 17:54

Let $$g(x) =f(x) - x$$ so that $$g'(x) \to L<0$$ as $$x\to\infty$$ and therefore $$g'(x)$$ is negative as $$x\to\infty$$. It thus follows that $$g(x)$$ eventually decreases and hence either tends to a limit $$M$$ or to $$-\infty$$. If it tends to $$-\infty$$ then $$g$$ is eventually negative and we are done. But if $$g(x) \to M$$ then by mean value theorem $$g'(\xi) = g(x+1)-g(x)\to 0$$ and this contradicts that $$g'(x) \to L<0$$.

Note: There is no need to assume that $$f$$ is continuously differentiable and we just need the hypotheses that $$\lim_{x\to\infty} f'(x)$$ exists and this limit is less than $$1$$.

• Paramand.Very nice and clear, as usual +. – Peter Szilas Oct 4 '18 at 17:53

So I think it is safe to say that the hypothesis

$$\displaystyle \lim_{x \to \infty} f'(x) < 1 \tag 1$$

gives us two important facts concerning the function $$f(x)$$:

first, that $$\lim_{x \to \infty} f'(x)$$ actually exists, that is, there is some $$L \in \Bbb R$$ such that, given any $$\epsilon > 0$$, there also exists $$M_\epsilon \in \Bbb R$$ such that

$$x \ge M_\epsilon \Longrightarrow \vert f'(x) - L \vert < \epsilon; \tag 2$$

second, that in fact

$$L < 1; \tag 3$$

then

$$1 - L > 0, \tag 4$$

so we can choose, say

$$\epsilon \le \dfrac{1 - L}{2} = \dfrac{1}{2}(1 - L); \tag 5$$

then we may write (2) in the form

$$x \ge M_\epsilon \Longrightarrow L - \epsilon < f'(x) < L + \epsilon \le L + \dfrac{1}{2}(1 - L) = \dfrac{1}{2}(L + 1) < 1; \tag 6$$

the key here is that by taking $$x$$ sufficiently large, we may ensure $$f'(x)$$ is bounded by a number, in this case $$(1/2)(L + 1)$$, which is strictly less than $$1$$ which, intuitively, shows us that $$f(x)$$ grows more slowly than $$x$$ for $$x$$ large enough.

Now consider the function $$x - f(x)$$; for $$\xi \ge M_\epsilon$$ we have

$$x - f(x) - (\xi - f(\xi)) = \displaystyle \int_\xi^x (s - f(s))' \; ds$$ $$=\displaystyle \int_\xi^x (1 - f'(s)) \; ds \ge \int_\xi^x \dfrac{1}{2}(1 - L) \; ds = \dfrac{1}{2}(1 - L)(x - \xi), \tag 7$$

or

$$x - f(x) \ge \dfrac{1}{2}(1 - L)(x - \xi) +(\xi - f(\xi)) \to \infty \; \text{as} \ x \to \infty \tag 8$$

since

$$\dfrac{1}{2}(1 - L) > 0, \tag 9$$

that is, the right-hand side of (8) describes a line of positive slope; we further see from (8) that this line intersects the $$x$$-axis at that unique $$x$$ such that

$$\dfrac{1}{2}(1 - L)(x - \xi) +(\xi - f(\xi)) = 0, \tag 9$$

i.e., where

$$x = -\dfrac{2(\xi - f(\xi))}{1 - L} + \xi; \tag{10}$$

therefore

$$x > f(x) \tag{11}$$

for

$$x > -\dfrac{2(\xi - f(\xi))}{1 - L} + \xi. \tag{12}$$

Nota Bene: It is not necessary to assume $$f(x)$$ actually has a limit as $$x \to \infty$$; scrutiny of the above argument reveals that the desired result, $$f(x) < x$$ for sufficiently large $$x$$, binds under the weaker assumption that

$$f'(x) \le L < 1, \; \forall x \; \text{"big enough"}. \tag{13}$$

End of Note.

• +1 for the note at the end. – Paramanand Singh Oct 5 '18 at 0:55