MHB What Happens to the Limit of This Function as T Approaches Infinity?

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The discussion centers on evaluating the limit of the function as T approaches infinity, specifically $$\lim_{{T}\to{\infty}}N \bar{h}\omega \left( \frac{1}{2} + \frac{1}{e^{\frac{ \bar{h}\omega}{k_BT}}-1} \right)$$. It is noted that as T increases, the term $$\frac{1}{e^{\frac{ \bar{h}\omega}{k_BT}}-1}$$ approaches an indeterminate form of $$\frac{1}{0}$$, suggesting the need for further analysis. The concept of an oblique asymptote is introduced, indicating that the limit will not stabilize but rather grow linearly as T increases, leading to $$E\to Nk_BT$$. An example function is provided to illustrate the concept of oblique asymptotes and their slopes. The discussion emphasizes the importance of understanding how the function behaves as T approaches infinity.
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$$\lim_{{T}\to{\infty}}N \bar{h}\omega \left( \frac{1}{2} + \frac{1}{e^{\frac{ \bar{h}\omega}{k_BT}}-1} \right)$$
In term $$\lim_{{T}\to{\infty}}N \bar{h}\omega \left( \frac{1}{e^{\frac{ \bar{h}\omega}{k_BT}}-1} \right)=N \bar{h}\omega \lim_{{T}\to{\infty}}\left( \frac{1}{e^{\frac{ \bar{h}\omega}{k_BT}}-1} \right)$$
but $$\left( \frac{1}{e^{\frac{ \bar{h}\omega}{k_BT}}-1} \right)$$ in the form $$\frac{1}{0}$$
please give me a idea

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Hi Another,

The expression for $E$ still has a $T$ in it when $T\to\infty$.
It means that we are looking for an oblique asymptote instead of a horizontal asymptote.
That is, if $T\to\infty$ we should find that $\pd E T\to Nk_B$, so that that we get the oblique asymptote $E\to Nk_BT$.
 
Klaas van Aarsen said:
Hi Another,

The expression for $E$ still has a $T$ in it when $T\to\infty$.
It means that we are looking for an oblique asymptote instead of a horizontal asymptote.
That is, if $T\to\infty$ we should find that $\pd E T\to Nk_B$, so that that we get the oblique asymptote $E\to Nk_BT$.

I do not understand, please give an example to explain it?

why $\pd E T\to Nk_B$, ?
 
Another said:
I do not understand, please give an example to explain it?

why $\pd E T\to Nk_B$, ?

Consider the function given by $f(x)=\frac 1x + x + 1$.
It has:
$$\lim_{x\to\infty} f(x) = \infty $$

[DESMOS]advanced: {"version":7,"graph":{"squareAxes":false,"viewport":{"xmin":-10,"ymin":-10,"xmax":10,"ymax":10}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"y=\\frac{1}{x}+x+1"},{"type":"expression","id":"2","color":"#388c46","latex":"y=x+1"}]}}[/DESMOS]
It also has the oblique asymptote $y=x+1$.
We can find its slope through the limit of $f'(x)$ when $x\to\infty$.
We have:
$$\lim_{x\to\infty} f'(x) =\lim_{x\to\infty} \left(-\frac{1}{x^2} + 1\right) = 1$$
Therefore the oblique asymptote exists and has slope $1$.
Put otherwise, when $x\to\infty$ we have that $f(x)\to x$.
Note that we didn't find the $y$-intercept yet, which we didn't need.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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