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archaic
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Would this ##f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0## hold if ##\frac{f(x)}{g(x)}\neq c##?
archaic said:Would this ##f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0## hold if ##\frac{f(x)}{g(x)}\neq c##?
Math_QED said:It is unclear to me what you mean. Can you be a little more careful describe what you mean? You have an ##\implies## and then an 'if' after the implies.
What is ##a## ? What is ##c##? Where are ##f,g## defined?
What I want to say is, if there're two functions ##f(x)## and ##g(x)## who are going to ##0## as ##x\to\infty##, and such that after a certain ##x\,(=a)## we have ##f'(x)>g'(x)## then ##\frac{f(x)}{g(x)} \underset{\infty}{\to} 0## provided that ##f(x)\neq cg(x)## where ##c\in\mathbb{R}##, elsewise the limit would equal ##c##.RPinPA said:This is a little cryptic. Here's what I think you are saying:
"There is a theorem that if for some ##a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)## and ##f,\,g\underset{\infty}{\to}0## (Does this mean ##\lim_{x\to\infty}f, g = 0##?) then ##\lim_{x\to\infty}f(x)/g(x)=0##. Does the theorem hold if ##\frac{f(x)}{g(x)}\neq c##?"
Or are you asking if there is such a theorem?
No, they don't come under this since they have the same rate of change.WWGD said:If I understood correctly, ##1/x ,1/(x+1)## are counters.
Respective derivatives are ##-1/x^2 >-1/(x+1)^2##archaic said:No, they don't come under this since they have the same rate of change.
Right sorry. I think, then, that I can say something about the convergence of the limit given the assumptions? And then it follows that it is ##0## if the functions are not equivalent at ##\infty##?WWGD said:Respective derivatives are ##-1/x^2 >-1/(x+1)^2##
archaic said:What I want to say is, if there're two functions ##f(x)## and ##g(x)## who are going to ##0## as ##x\to\infty##, and such that after a certain ##x\,(=a)## we have ##f'(x)>g'(x)## then ##\frac{f(x)}{g(x)} \underset{\infty}{\to} 0## provided that ##f(x)\neq cg(x)## where ##c\in\mathbb{R}##, elsewise the limit would equal ##c##.
I realized some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.
However, since what I want to say is if ##f## goes to ##0## faster than ##g## then ..., the correct way to translate this in math is ##|f'(x)|<|g'(x)|## (provided that both ##f## and ##g## are going to ##0##).
$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$PeroK said:I can't see any reason why this would be true.
archaic said:$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$
By $10^{k}$ I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use ##O(10^{-n})##? Not sure, I need to study big-O notation.
Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.PeroK said:I can't see the relationship between that and what you posted before.
archaic said:Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.
archaic said:Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.
If ##f## is going faster to ##0## than ##g##, then ##|df|## would be less than ##|dg|##, because it is already "at zero" so there isn't much to change.PeroK said:That relationship means that ##g(x)## is changing more quickly than ##f(x)##.
You've got that the wrong way round.
Right, see post #8.PeroK said:In any case, you already have the counterexample: ##f(x) = \frac 1 2 g(x)##.
Even if you exclude that precise counterexample, you only need two functions where ##f(x) \approx \frac 1 2 g(x)##. For example:
##f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)##
archaic said:however, given the premises, it seems that the limit always converges (maybe), and in the special case that ##f\underset{\infty}{\not\sim}g##, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.
wopsPeroK said:Not that either. Just amend my last function:
##f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)##
Then the limit ##f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}## is undefined.
It doesn't satisfy ##|f'(x)|<|g'(x)|## so it doesn't come under this.PeroK said:Not that either. Just amend my last function:
##f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)##
Then the limit ##f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}## is undefined.
archaic said:wops
With ##g(x) = e^{-x}## it all works out.archaic said:It doesn't satisfy ##|f'(x)|<|g'(x)|## so it doesn't come under this.
you mean given the premises?PeroK said:Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.
The only one we don't have is where ##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty##.
Any ideas?
Yes.archaic said:you mean given the premises?
archaic said:you mean given the premises?
I thought you meant that there was an example where that limit goes to infinity...PeroK said:Maybe this is the result that could be proved? If ##|f'(x)| < |g'(x)|##, then ##f(x)## is changing less than ##g(x)##, so ##f(x)/g(x)## cannot become arbitrarily large. So:
##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty##?
... looking for counterexamples is good way to see what might be true.archaic said:I thought you meant that there was an example where that limit goes to infinity...
Hello again PeroK. Another way to put ##|f'(x)|<|g'(x)|## would be ##|f(x)|'>|g(x)|'## (for oscillating functions only in intervals where they are both positive, but the last step (the squaring) fixes this). Taking that into account, we havePeroK said:... looking for counterexamples is good way to see what might be true.
A limit in mathematics is a fundamental concept that describes the behavior of a function as its input approaches a certain value. It is used to determine the value that a function approaches as its input gets closer and closer to a specific value, without actually reaching it.
Limits are important in science and engineering because they allow us to make predictions and analyze the behavior of systems. They are used to model real-world phenomena and make calculations that would otherwise be impossible.
A one-sided limit only considers the behavior of a function as its input approaches the specific value from one direction, either from the left or the right. A two-sided limit considers the behavior from both directions and requires the function to approach the same value from both sides.
To find the limit of a function, you can either use algebraic techniques or graphing techniques. Algebraically, you can evaluate the function at values approaching the specific value and see what value the function approaches. Graphically, you can plot the function and visually determine the behavior of the function as its input approaches the specific value.
Limits have many real-world applications, such as calculating the maximum safe speed of a vehicle on a curved road, determining the maximum load a bridge can hold, and predicting the population growth of a species. They are also used in fields such as economics, physics, and engineering to model and analyze various systems and phenomena.