What is $\lim\limits_{b\to a}\frac{e^{-\frac{x}{a}}-e^{-\frac{x}{b}}}{a-b}$?

Question

I'm trying to evaluate $$\lim\limits_{b\to a}\frac{e^{-\frac{x}{a}}-e^{-\frac{x}{b}}}{a-b}$$

I know that the limit exists. The limit of the numerator and denominator are both zero when $b\to a$, so I tried to apply L'Hospital's Rule for $\frac00$ form, but the denominator is constant with respect to $x$ so it becomes zero and then the whole thing is undefined.

How does one go about tackling this limit?

Answer 1

This is nothing but

$$\frac{d}{da}e^{-\frac xa} = \frac x{a^2}e^{-\frac xa}$$

To see this, note that

$$\lim\limits_{b\to a}\frac{e^{-\frac{x}{a}}-e^{-\frac{x}{b}}}{a-b} = \lim\limits_{b\to a}\frac{e^{-\frac{x}{b}}-e^{-\frac{x}{a}}}{b-a}$$

Answer 2

Apply L'Hospital rule for $\frac00$ form & differentiate w.r.t. $b$ (where $b$ is taken as variable for the limit while $x$ & $a$ are constants) as follows $$\lim\limits_{b\to a}\frac{e^{-\frac{x}{a}}-e^{-\frac{x}{b}}}{a-b}=\lim\limits_{b\to a}\frac{\frac{d}{db}\left(e^{-\frac{x}{a}}-e^{-\frac{x}{b}}\right)}{\frac{d}{db}(a-b)}$$ $$=\lim\limits_{b\to a}\frac{0-e^{-\frac{x}{b}}\frac{x}{b^2}}{-1}$$ $$=\lim\limits_{b\to a}\frac{x}{b^2}\cdot e^{-\frac{x}{b}} $$ $$=\frac{x}{a^2}e^{-\frac{x}{a}}$$

Answer 3

Let $b = a+h$. then, for small $h$,

$\begin{array}\\ \dfrac{e^{-\frac{x}{a}}-e^{-\frac{x}{b}}}{a-b} &=\dfrac{e^{-\frac{x}{a}}-e^{-\frac{x}{a+h}}}{-h}\\ &=\dfrac{e^{-\frac{x}{a}}(1-e^{-\frac{x}{a+h}+\frac{x}{a}})}{-h}\\ &=\dfrac{e^{-\frac{x}{a}}(1-e^{-\frac{x}{a+h}(1-\frac{a+h}{a})})}{-h}\\ &=\dfrac{e^{-\frac{x}{a}}(1-e^{-\frac{x}{a+h}(\frac{a-(a+h)}{a})})}{-h}\\ &=\dfrac{e^{-\frac{x}{a}}(1-e^{\frac{xh}{a(a+h)}})}{-h}\\ &\approx\dfrac{e^{-\frac{x}{a}}(1-e^{\frac{xh}{a^2}})}{-h}\\ &\approx\dfrac{e^{-\frac{x}{a}}(1-(1+\frac{xh}{a^2}))}{-h}\\ &\approx\dfrac{-e^{\frac{x}{a}}\frac{xh}{a^2}}{-h}\\ &=\dfrac{xe^{-\frac{x}{a}}}{a^2}\\ \end{array} $