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Say I have a function $f\left(x\right)$ which does not have a limit as $x \to \infty$. My question is whether there is a method for finding another function $g\left(x\right)$ that the function converges to in the sense that the difference $f - g$ goes to $0$ as $x \to \infty$.

I am aware that there may be many such functions, so I am not expecting such a method to give all such functions, but hopefully at least one function $g\left(x\right)$ which is simpler to work with than $f$.

I'll give an example to convey my motivation here. Consider the function $f = {{1}\over{\exp{\left(1\over{x}\right)-1}}}$. If I want to know the behaviour as $x \to \infty$ and I expand the exponential in a power series and just keep the first term I get $f = {{1}\over{1 + {1\over{x}} - 1}}=x$, but when I look at the graph of $f$ it seems to actually converge to the function $y = x - {1\over2}$. I am wondering how I could find that out algebraically (by which I mean without looking at the graph). This is not about showing that $f$ in this example converges to $y = x - {1\over2}$; I am looking for a way by which I could find that function $y$ in the first place.

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As in many questions, the answer is Bernoulli, but in this case the numbers, not the inequality. See https://en.wikipedia.org/wiki/Bernoulli_number.

You have $f(x) = {{1}\over{\exp{\left(1\over{x}\right)-1}}} $.

The Bernoulli numbers satisfy $b(t) =\dfrac{t}{e^t-1} =\sum_{m=0}^{\infty} \dfrac{B_m t^m}{m!} $ where the $B_m= 1, -\dfrac12, \dfrac16, 0, -\dfrac1{30}, ... $.

Therefore $b(1/x) =\dfrac{1/x}{e^{1/x}-1} $ so that $f(x) =xb(1/x) =x(1 -\dfrac1{2x}+ \dfrac1{6x^2} -\dfrac1{30x^4}+ ...) =x -\dfrac1{2}+ \dfrac1{6x} -\dfrac1{30x^3}+ ... $.

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Linear asymptotes may be deduced as follows (if there are linear asymptotes)

$$m=\lim_{x\to\infty}\frac{f(x)}x$$

$$b=\lim_{x\to\infty}f(x)-mx$$

Thus, the linear asymptote is then given by

$$f(x)\sim mx+b$$

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$e^{1/x}-1 =1/x+1/(2x^2)+... =(1/x)(1+1/(2x)+...) $

so

$1/(e^{1/x}-1) =x/(1+1/(2x)+...) =x(1-1/(2x)+...) =x-1/2+... $

Take more terms to get more accuracy.

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