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I am trying to figure out why this is true:

$$ \lim_{p \to 0}\frac{1}{2p}\left((1+p)e^{-\frac{y}{1+p}} - (1-p)e^{-\frac{y}{1-p}}\right) = e^{-y} + ye^{-y}$$

I have already tried L'Hopital's Rule, but it gave me something that I couldn't simplify. The problem seems to be the $\frac{1}{2p}$ term never seems to go away. I know the exponential function can be represented as: $e^x = \lim_{n\to\infty} (1+\frac{x}{n})^n$, but it doesn't seem immediately obvious how that would apply in this situation.


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L'Hôpital is the right approach and a single application yields the answer you give. Did you remember to set $p=0$ after taking the derivative of the numerator? –  joriki Apr 3 '11 at 20:07
Hi Joriki, I did use L'Hopital, and I got a very unfriendly function, and I did set p to zero. I'm using the product rule. First term is $\frac{1}{2p}$ and second term is the rest. So the first term still exists after doing the derivative (unless something cancelled out, but I could not see how the 2p would cancel out). –  jrand Apr 3 '11 at 20:13
In the application of L'Hospital's rule, you should take the derivative of the numerator and the derivative of the denominator of the function separately. You seemed to have taken the derivative of the entire function. –  JavaMan Apr 3 '11 at 20:15
I think that's a misunderstanding of how L'Hôpital applies to this. $2p$ is the denominator; you need to differentiate it separately, not as part of a product. The rule you need is that if $\lim a$ and $\lim b$ vanish, then $\lim \frac{a}{b}$ is $\lim\frac{a'}{b'}$, where $b=2p$. –  joriki Apr 3 '11 at 20:18
@joriki, DJC: Yes, after I took the derivatives separately and set p to zero, I got the correct result. Thanks. –  jrand Apr 3 '11 at 20:23

4 Answers 4

up vote 5 down vote accepted

HINT $\ $ It has form $\displaystyle\rm\ \lim_{x\to 0}\ \frac{f(x)-f(-x)}{x - (-x)}\:.\ $ Relate that to a derivative.

Note that this solution by recognizing the limit as a derivative employs only knowledge of the definition of the derivative and the basic rules for calculating derivatives of polynomials and powers. It does not require knowledge of more advanced techniques such as power series or Taylor series, l'Hôpital's rule, the mean-value theorem, etc. As I often remark here, limits exercises are frequently of this form (e.g. see this prior question or this or this or this or this) so it is worthwhile to be familiar with this technique, which - when it applies - is often much simpler than alternatives.

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I tried this and it worked, although I couldn't get it in the above form exactly. $f(p) = (1+p)e^{-\frac{y}{1+p}} - (1-p)e^{-\frac{y}{1-p}}$. $f(0) = 0$. Then, the original limit is equal to: $\frac{1}{2} \lim_{p \to 0} \frac{f(p) - f(0)}{p} = \frac{1}{2} f'(0)$ –  jrand Apr 3 '11 at 20:51

Hint: Writing the numerator of the fraction in the limit as

$$ p \left( e^{- \frac{y}{1+p}} + e^{- \frac{y}{1-p}} \right) + e^{- \frac{y}{1+p}} - e^{- \frac{y}{1-p}} $$

should help.

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I took the derivatives of top and bottom, and it gave me the correct result. This method requires me to take the derivative of $e^{\frac{-y}{1 \pm p}}$ instead of $p e^{\frac{-y}{1 \pm p}}$. –  jrand Apr 3 '11 at 20:33

Here is one approach that does not depend on recognizing the expression as a derivative, and allows you to use L'Hôpital's rule; I write $\exp(x)$ for $e^x$ for clarity:

You can write $\displaystyle \exp\left(-\frac{y}{1+p}\right)-\exp\left(-\frac{y}{1-p}\right)$ as $$\exp\left(-\frac{y}{1-p^2}\right)[\exp\left(\frac{yp}{1-p^2}\right)-\exp\left(-\frac{yp}{1-p^2}\right)].$$

Now note that $\displaystyle \frac{(1+p)\exp\left(-\frac{y}{1+p}\right)-(1-p)\exp\left(-\frac{y}{1-p}\right)}{2p}=$ $$\frac{\exp\left(-\frac{y}{1+p}\right)-\exp\left(-\frac{y}{1-p}\right)}{2p}+\frac12(\exp\left(-\frac{y}{1+p}\right)+\exp\left(-\frac{y}{1-p}\right)).$$

The second term in this sum can be evaluated directly. For the first, note that $\displaystyle e^\alpha-e^{-\alpha}=\frac{e^{2\alpha}-1}{e^\alpha}$. In this case, $\displaystyle \alpha=\frac{yp}{1-p^2}$, so the problem reduces to finding $$ \lim_{p\to 0}\frac{e^{\frac{2yp}{1-p^2}}-1}{2p},$$ as all the other terms can be evaluated directly. But this term can now be evaluated without difficulties using L'Hôpital's rule.

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@Bill Dubuque: I removed your comment as it has nothing to do with this answer. Please use e-mail or similar methods of communication for that purpose. –  Willie Wong Apr 6 '11 at 16:09

In fact I don't recommend always trying L'Hopital's rule on every limit you come across, because it often fails and rarely gives an intuitive insight into the limit itself. Here is an alternative method that essentially obtains an asymptotic expansion, immediately making the limiting behaviour apparent.

As $p \to 0$:

  $\frac{1}{2p}\left((1+p)e^{-\frac{y}{1+p}} - (1-p)e^{-\frac{y}{1-p}}\right)$

  $ \in \frac{1}{2p}\left((1+p)e^{-y(1-p+Θ(p^2))} - (1-p)e^{-y(1+p+Θ(p^2))}\right)$

  $ ⊆ \frac{1}{2p}\left((1+p)e^{-y}e^{yp+Θ(p^2)} - (1-p)e^{-y}e^{-yp+Θ(p^2)}\right)$

  $ ⊆ e^{-y}\frac{1}{2p}\left((1+p)(1+yp+Θ(p^2)) - (1-p)(1-yp+Θ(p^2))\right)$

  $ ⊆ e^{-y}\frac{1}{2p}\left((1+(y+1)p+Θ(p^2)) - (1-(y+1)p+Θ(p^2))\right)$

  $ ⊆ e^{-y}\frac{1}{2p}\left(2(y+1)p+Θ(p^2)\right)$

  $ ⊆ e^{-y}(y+1)+Θ(p)$

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