Stability?

Calculus Level 3

If function $f(x)$ satisfies $\lim_{x\to \infty}(f(x)+f'(x))=C, C\in \mathbb{R}, \text{and }C\ne 0$ Evaluate $\lim_{x\to \infty}f(x)$

It is divergent

0

C

Information given is not sufficient

2 solutions

Otto Bretscher
Nov 23, 2018

We can use Bernoulli's Rule (often misattributed to de l'Hôpital): $\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{e^xf(x)}{e^x}=\lim_{x\to \infty}\frac{(e^xf(x))'}{(e^x)'}=\lim_{x\to \infty}\frac{e^xf(x)+e^xf'(x)}{e^x}=\lim_{x\to \infty}(f(x)+f'(x))=\boxed{C}$ .

The condition $C \neq 0$ is unnecessary, I believe.

But how do you know $\lim_{x\to \infty}{e^xf(x)}=\infty$ . Otherwise you can't use Bernoulli's Rule.

Swafim Li - 2 years, 6 months ago

Sure I can use it. Bernoulli's rule only requires that the denominator go to infinity. See Theorem 5.13 in "Baby Rudin," for example. (In introductory calculus texts, these issues are usually not handled well.)

Otto Bretscher - 2 years, 6 months ago

You let me learn something new. Thanks

Swafim Li - 2 years, 6 months ago

A Former Brilliant Member
Nov 26, 2018

I think we can clearly say that $f(x) = e^{-x} + C$

As, $f(x) + f'(x) = e^{-x} + C + (-e^{-x} + 0) = C$

Hence, $\displaystyle \lim_{x\rightarrow \infty} (f(x) + f'(x)) =\displaystyle \lim_{x\rightarrow \infty} C = C$

Hence, we get $\displaystyle \lim_{x\rightarrow \infty} f(x) = \displaystyle \lim_{x\rightarrow \infty} (e^{-x} + C )= e^{-\infty} + C = 0+ C = \boxed{C}$

Stability?

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