Mathematical Analysis?

Question

f: R -> R be a differetiable function. Assume that lim(x->infinity) f(x) = a for some real number a, compute lim(x->infinity) x * f'(x) Can someone help me with this problem. Show me your work. Thanks!

Answer 1

There isn't enough information to determine whether this limit exists. Consider the function f(x) defined by f(x) = {sin x/x + a if x≥π, -(x-π)/π + a if x<π} (the piecewise definition is absolutely unimportant, I just put it there to avoid the obvious singularity at zero so that the function is differentiable on all of R. The important part is that past a certain point, the function is sin x/x + a). Clearly, [x→∞]lim f(x) = a, and f'(x) = (x cos x - sin x)/x² (for x≥π), so x*f'(x) = cos x - sin x/x (for x≥π), and [x→∞]lim x*f'(x) does not exist, because x*f'(x) oscillates between 1 and -1. Thus this limit does not always exist. However, if I had used instead f(x) = {sin x/x² + a if x≥π, -(x-π)/π² + a if x<π}, then [x→∞]lim x*f'(x) would exist, and would equal zero.

Although we cannot determine, just from what is written, whether or not the limit in question exists, we can show that IF it exists, it must be 0. Consider the following:

Suppose [x→∞]lim x*f'(x) = c, where c>0. Then this means that for all ε>0, there exists an N such that for all x>N, x*f'(x) ∈ [c-ε, c+ε]. In particular, letting ε=1/2 we have that x*f'(x)∈[c/2, 3c/2] and thus x*f'(x) ≥ c/2. Without loss of generality we assume that N is positve, so x is also positve and we can divide by it without changing the inequality -- thus, f'(x) ≥ c/(2x) Now consider that f(x) = f(N) + [N, x]∫f'(t) dt by the fundamental theorem of calculus. Thus, for x>N, f(x) ≥ f(N) + [N, x]∫c/(2t) dt = f(N) + c/2 [N, x]∫1/t dt. However, [x→∞]lim f(N) + c/2 [N, x]∫1/t dt = ∞, so this means [x→∞]lim f(x) =∞, contradicting the assumption that it is a real number.

Conversely, if c<0, then we can show that f'(x) ≤ c/(2x), so f(x) ≤ f(N) + c/2 [N, x]∫1/t dt, and since c is negative, [x→∞]lim f(N) + c/2 [N, x]∫1/t dt = -∞, so [x→∞]lim f(x) = -∞, again contradicting the assumption that it is a real number.

So if [x→∞]lim x*f'(x) exists, it is 0.

Answer 2

So many big words ^_º

Well if x is approaching infinity and the result is just getting closer to a, you know that the limit of the slope (derivitave) as x approached infinity is 0.

I'd say 0.

No that's wrong... I know how to do this but I can't remember :-(

Oh well maybe I was right, James agrees with me :-)

Answer 3

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Answer 4

lim(x->infinity) x * f'(x) =
infinity or
0 or
(- )infinity

depending on whether
a>0 or
a=0 or
a<0.

Answer 5

f'(x) = 0 => x * f'(x) = 0
=> lim(x->infinity) x * f'(x) = 0

Answer 6

I vote for 0, but thus far have not been able to prove this to myself formally ... it seems like you can say f(x) = a + o(1), so f'(x) = o(1/n), so xf'(x) = o(1), so the limit is 0, but that does not seem completely formal ...

Answer 7

If f(x) - - > a when x - - >inf, then f(x) = a + a1 / x + a2 / x^2 +++ is tailor series; thence f’(x) = -a1/x^2 –2*a2/x^3 ---;
and xf’(x) = -a1/x –2*a2/x^2 ---;
Since k*aj are constants, then xf’(x) -- >0;