Does the $\lim_{x \to \infty} \frac{g(x)}{e^x}$ exist?

Question

Suppose for all real $x$ that $g^{(n)}(x) = g^{(n-1)}(x)$ and $g^{(n-1)}(0) = -1$ where $g^{n}(x)$ is the nth derivative. Does the $\lim_{x \to \infty} \frac{g(x)}{e^x}$ exist?

Isn't this answer just the form $g(x) = -e^x$? and Hence $\lim_{x \to \infty} \frac{g(x)}{e^x} = -1$? How could I more rigorously show this?

Is $n$ fixed or should $g$ be such that equalizes hold for arbitrary $n$? — mihaild, Nov 11 '20 at 17:21
This holds for one fixed $n$? I think $g(x)$ will be $-e^x$ plus a polynomial of degree $n-2$. Thus your limit is $-1$. — GEdgar, Nov 11 '20 at 22:27

score 1 · Answer 1 · answered Nov 11 '20 at 17:59

1

Weird. Isn't that the function $h(x):= e^{-x}g(x)$ has null derivative?
Which implies $h$ is a constant function.
Is there any other constraint?

answered Nov 11 '20 at 17:59

Paresseux Nguyen

7,331

Why is it weird? Also, nice catch that we don't need general uniqueness theorem here. – mihaild Nov 11 '20 at 18:02
Yeah. Because most of the time, for a similar problem, we can't calculate explicitly the formula of $g$. – Paresseux Nguyen Nov 11 '20 at 18:05
I mean it asks if the limit exists ... I suppose you could prove it does not as well, but the question is worded exactly as given. – A. Radek Martinez Nov 11 '20 at 19:23
I just confirmed your thought that $g(x)=-e^{x}$ – Paresseux Nguyen Nov 11 '20 at 19:27

score 1 · Answer 2 · answered Nov 11 '20 at 21:46

1

The taylor series of a function is given as: $$g(x)=\sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}x^n$$ so in your case: $$g(x)=-\sum_{n=0}^\infty\frac{x^n}{n!}=-e^x$$ so: $$\frac{g(x)}{e^x}=-1$$ which is not dependent on $x$ and so converges for all $x$

answered Nov 11 '20 at 21:46

Henry Lee

12,215

Doesn't that require to prove that $g$ if analytical? – mihaild Nov 12 '20 at 12:18

Does the $\lim_{x \to \infty} \frac{g(x)}{e^x}$ exist?

2 Answers2