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# Function bounds from derivative limits (TIFR 2014 problem 3)

Question:

Let $$f:\mathbb{R}\to \mathbb{R}$$ be a differentiable function such that $$\lim_{x\to \infty} f'(x)=1$$ then

A. f is bounded

B. f is increasing

C. f is unbounded

D. f’ is bounded

Discussion:

The easiest function to look at is the function which has derivative 1 at all points. That is $$f(x)=x$$. This function is not bounded. So option A is wrong.

Now, the function $$f$$ is increasing. But let’s say we disturb the same $$f$$ around 0 a little bit, so that the function is not increasing any more. We can do this disturbing so that the function still remains differentiable, but it loses the increasing part of it. This tells us that B is wrong.

We can disprove D in a similar manner, for even though $$f’$$ can be bounded in the positive side of x-axis, what if $$f’\to \infty$$ as $$x\to -\infty$$? And this case in absolutely possible, because we have no information about the nature of $$f$$ for -ve values of x. We can construct such an example easily by the help of drawing graphs, or constructing functions explicitly. (Use $$x^2$$ when x is negative and some linear function/translation of x when x is positive, and make sure the function transition is smooth in between).

Now, we have eliminated all but option C. We prove option C.

What does the expression (\lim_{x\to \infty} f'(x)=1\) mean?

It means that for large values of $$x$$ $$f'(x)$$ is close enough to 1.

That means, for a sequence going to $$\infty$$ we have $$f’$$ at the sequence-points going to 1.

Take the sequence as $$n\in \mathbb{N}$$. Then

$$\lim_{n\to \infty} lim_{h \to 0} \frac{f(n+h)-f(n)}{h} =1$$.

Let’s forget about $$h\to 0$$. Let’s say, h=1.

Then $$f(n+1)-f(n)=f'(y_n)(n+1-n)$$ by the mean-value theorem. Where $$y_n\in[n,n+1]$$. As $$n\to \infty$$, $$y_n \to \lim_{x\to \infty} f'(x)=1$$.

That means, given $$\epsilon >0$$; for large values of $$n$$

$$|f(n+1)-f(n)-1|=|f'(y_n)-1|< \epsilon$$. This means that $$f$$ increases approximately by 1 for an increment of 1 in the value of $$x$$. This means that $$f$$ behaves like $$x$$ for large values of $$x$$. Which proves that $$f$$ must be unbounded.

September 30, 2017