Limit involving exponential functions 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.
Thanks.
 A: 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.
A: 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.
A: 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.
A: 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)$
