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I'm solving a problem about a plug flow reactor and I have this limit to compute. Just to control my result I asked Wolfram and I'm confuse can you explain me the result please.

I precise $x$ is a fixed value.

$$\lim_{R \to +\infty} \frac{1-\exp\left(\frac{x}{R+1}\right)}{\frac{R}{R+1}-\exp\left(\frac{x}{R+1}\right)}$$

When I made my reasonning I said, when $R$ goes to the infinity then the exponential terms both go to zero then the limit is the limit of $(1+R)/R$ which goes to $1$ when $R$ goes to the infinity.

So I bet my reasonning is false but don't know why. Please don't answer me with the L'Hopital theorem I dislike it.

Thank you in advance for your answer.

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    $\begingroup$ The exponential terms tend to $\exp(0) = 1$ when $R\to \infty$. Thus you get an indeterminate form, and the limit depends on how fast numerator and denominator each approach $0$. The numerator is $-\frac{x}{R+1} + O(R^{-2})$, and the denominator $-\frac{x+1}{R+1} + O(R^{-2})$ by Taylor expansion. $\endgroup$ – Daniel Fischer Dec 22 '16 at 16:21
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Set $x/(R+1)=t$, so $R+1=x/t$ and $$ \frac{R}{R+1}=\frac{x/t-1}{x/t}=\frac{x-t}{x} $$ Then you have, depending on whether $x>0$ or $x<0$, the limit for $t\to0^+$ or the limit for $t\to0^-$. Let's compute the two-sided limit: $$ \lim_{t\to0}\frac{1-e^t}{\frac{x-t}{x}-e^t}= \lim_{t\to0}x\frac{1-e^t}{x-t-xe^t}= \lim_{t\to0}x\frac{1-1-t+o(t)}{x-t-x-xt+o(t)}=\frac{x}{1+x} $$ For $x=0$, the limit poses no problem and is $0$, so the formula is valid for $x\ne-1$.

For $x=-1$, we have $$ \lim_{t\to0^{-}}\frac{1-e^t}{1+t-e^t}=-\infty $$

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You cannot consider different expressions depending on the same variable one by one!

With your same reason the well known sequence $\left(1 + \frac{1}{n}\right)^n$ would converge to $1$ because the expression in the brackets tends to 1 and $1^n = 1$. But this is wrong and $$\lim_{n\to \infty} \left(1 + \frac{1}{n}\right)^n = e$$ holds.

And so you cannot consider each expression depending on $R$ by it's own.

Also you cannot use limit rules here which leads to an "$\frac{0}{0}$" result, and although you dislike L'Hopital theorem it should be the theorem of choice here…

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  • $\begingroup$ I'm yeah I was thinking about but not sure. OK so I will learn this theorem, I've never seen it in any of my lessons. It will be quite fun. Thank you. $\endgroup$ – ParaH2 Dec 22 '16 at 16:23
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Use the fundamental limit:

$$\lim_{x\rightarrow \infty}x(e^{1/x}-1)=1 \Rightarrow \lim_{x\rightarrow \infty}x(e^{a/x}-1)=a$$

and write your limit like:

$$\lim_{R \rightarrow \infty}\left(\frac{(R+1)(1-e^{x/(R+1)})}{-1+(R+1)(1-e^{x/(R+1)})}\right)=\frac{x}{1+x}$$

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One has $$ \exp(h)=1+h+h\epsilon(h) $$ with $\lim_{h\rightarrow 0} \epsilon(h)=0$ then, as $R\rightarrow +\infty$ one has $$ \exp\left(\frac{x}{R+1}\right)=1+\frac{x}{R+1}+\frac{x}{R+1}\epsilon_1(R) $$ with $\lim_{R\rightarrow +\infty} \epsilon_1(R)=0$. Substituting this (exact !) expression in your quotient gives the result.

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You should use L'Hopital theorem. That gives us:

$$\lim_{R\rightarrow+\infty} = \frac{\frac{d}{dR}\bigl( 1 - \exp(\frac{x}{R+1}) \bigr)}{\frac{d}{dR}\bigl( \frac{R}{R+1} - \exp(\frac{x}{R+1} \bigr)}$$

This is equal to:

$$\lim_{R\rightarrow+\infty} \frac{\frac{x}{(R+1)^2}}{ \frac{R+1-R}{(R+1)^2}+\frac{x}{(R+1)^2}} = \frac{x}{x+1}$$

I'm not 100% sure about this, but I think that you must use L'Hopital theorem if you have indeterminate form, like $\frac{0}{0}$ etc.

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  • $\begingroup$ I don't know the hypothesis to use it and I have not time to verify them when I'm in an exam that's why I prefer not to use it. But thank's for your answer. $\endgroup$ – ParaH2 Dec 22 '16 at 16:18
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    $\begingroup$ the hypothesis for the most problems in chemistry is a form of "$\frac{0}{0}$" or "$\frac{\infty}{\infty}$". That's all you have to check and almost all student are able to do so. So you will be also ;-) $\endgroup$ – Gono Dec 22 '16 at 16:22
  • $\begingroup$ Yes I know but I'm like everybody the less I do the better I am ^^ $\endgroup$ – ParaH2 Dec 22 '16 at 16:24
  • $\begingroup$ L'Hopital theorem is really easy one to use. If you get an indeterminate form, then you just find a derivative of a nominator and denominator separately, and then try again with $\lim$. If you again get an indeterminate form, then find derivative again. Repeat this until you get a determinate form. Easy as that. $\endgroup$ – Marko Gulin Dec 22 '16 at 16:26

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