majmunomog

2022-09-03

$\underset{p\to 0}{lim}\frac{1}{2p}\left(\left(1+p\right){e}^{-\frac{y}{1+p}}-\left(1-p\right){e}^{-\frac{y}{1-p}}\right)={e}^{-y}+y{e}^{-y}$
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: ${e}^{x}=\underset{n\to \mathrm{\infty }}{lim}\left(1+\frac{x}{n}{\right)}^{n}$, but it doesn't seem immediately obvious how that would apply in this situation.

### Answer & Explanation

Shania Delacruz

HINT
It has form . 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.

bolton8l

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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