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I've been stuck on a certain HW problem for awhile now, and I'm not really sure how to go about solving it.

We're given the function:

$$f(x)=\frac{8\pi hcx^{-5}}{e^{hc/(xkT)}-1}$$

According to the problem, $h$, $c$, $k$, and $T$ are all defined as constants.

Use L'Hospital's Rule to show that both:

$$\lim _{x \to \ 0^+} f(x)=0$$ $$\lim _{x \to \ \infty} f(x)=0$$

My attempt for $\lim _{x \to \ \infty}$

$$\lim _{x \to \ \infty} f(x)=\frac{8\pi hcx^{-5}}{e^{hc/(xkT)}-1}$$

$$8\pi hc \lim _{x \to \ \infty} f(x)=\frac{\frac{1}{x^5}}{e^{hc/(xkT)}-1}$$

$$8\pi hc \lim _{x \to \ \infty} f(x)=\frac{\frac{1}{\infty}}{e^{hc/\infty}-1}=\frac{0}{0}$$

I then apply L'Hospital's Rule.

$$8\pi hc \lim _{x \to \ \infty} f(x)=\frac{\frac{-5}{x^6}}{\frac{-hce^{hc/(xkT)}}{kTx^2}}$$

I attempt to work on it so I could evaluate the limit.

$$8\pi hc \lim _{x \to \ \infty} f(x)=\frac{\frac{-5}{e^{hc/kTx}}}{\frac{hcx^4}{kT}}$$

I evaluate the limit.

$$\lim _{x \to \ \infty} f(x)=\frac{-5}{e^{hc/kTx}}=-5$$

$$\lim _{x \to \ \infty} f(x)=\frac{hcx^4}{kt}=\infty$$

$$\frac{-5}{\infty}=0$$

$$8\pi hc \times 0 = 0$$

Was what I did mathematically sound? Were there any errors? For the $\lim _{x \to \ 0^+}$, I gave it a go, but I couldn't think of a way to find good derivatives after my first one.

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closed as unclear what you're asking by Jack's wasted life, Jean Marie, user91500, Watson, Siminore Nov 4 '16 at 12:22

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    $\begingroup$ You don't show what is you real problem. Where are you stucked ? Have you tried to compute the derivatives of the numerator and of the denominator ? $\endgroup$ – Jean Marie Nov 3 '16 at 7:27
  • $\begingroup$ I've tried giving it a go and messing around with some algebra to make it work with L'Hospital's Rule. I think I have the correct answer for finding the limit as x approaches infinity. That required only one application of L'Hospital's rule. But I honestly have no idea where to start on finding the limit as x approaches 0. Maybe I'm just overlooking something simple? I've been chewing on this question for several days now. $\endgroup$ – Bilbo Nov 3 '16 at 7:36
  • $\begingroup$ Are your derivatives the good ones ? You should display them. Maybe the problem is there... $\endgroup$ – Jean Marie Nov 3 '16 at 7:41
  • $\begingroup$ I have checked: your computations are correct. Personally, I would have changed at once $y=1/x$ in the initial expression. It would have been a little simpler and less prone to errors... $\endgroup$ – Jean Marie Nov 3 '16 at 9:04
  • $\begingroup$ Ah, that's good then. Do you have any suggestions on how do the the limit as x approaches 0? I've been reworking the equation to use L'Hospital's rule, but what I end up with is either unnecessarily hard to derive, an indeterminate form, or a constant divided by 0. I don't know if I'm overthinking it or if I'm just missing something simple that I could've done. $\endgroup$ – Bilbo Nov 3 '16 at 9:14
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Knowing that $$f(x)=\frac{8\pi hcx^{-5}}{e^{hc/(xkT)}-1}$$

Let $a:=8\pi hc$, $b:=hc/(kT)$ and $y=1/x$. Then:

$$f(x)=g(y):=\frac{ay^{5}}{e^{by}-1}$$

First case :

$$\tag{1}\lim _{x \to \ \infty} f(x)=\lim _{y \to \ 0_+} g(y)=\lim _{y \to \ 0_+}\frac{ay^{5}}{by}=\lim _{y \to \ 0_+}\frac{a}{b}y^{4}=0$$

(I have replaced $e^{by}-1$ by an equivalent expression $by$ : it is classical that $\dfrac{e^{u}-1}{u} \equiv 1$ when $u \rightarrow 0$).

Remark: you could obtain the last equality in (1) by using L'Hospital too.

Second case :

$$\tag{2}\lim _{x \to \ 0_+} f(x)=\lim _{y \to \ \infty} g(y)=\lim _{y \to \ \infty}\frac{ay^{5}e^{-by}}{({e^{by}-1})e^{-by}}=\lim _{y \to \ 0_+} ay^{5}e^{-by} \frac{1}{1-e^{-by}}=0_+ \times 1=0_+$$

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