Limit applied on a function

Question

How do you determine the limit: $$\frac{e^{-ax} - e^{-bx}}{b-a}$$ as $b$ tends to $a$. I've tried using L'hospitals by couldn't get the right answer.... Any help would be appreciated

Answer 1

One way to solve it is to compare the expression with the definition of differentiation. For fixed $x$, define $f(t) = e^{-xt}$, then $$\frac{e^{-ax} - e^{-bx}}{b - a} = -\frac{f(b) - f(a)}{b - a} \to - f'(a)$$ as $b \to a$. Clearly, by the chain rule, $$f'(a) = -xe^{-xa}.$$ Hence the result is $xe^{-xa}$.

Answer 2

Rewrite $$A=\frac{e^{-ax} -e^{-bx} }{b-a}=e^{-ax}\frac{1-e^{(a-b)x} }{b-a}$$ Now, remember that, for small $y$, $e^y=1+y+\frac {y^2}2+\cdots$.So $$e^{(a-b)x}\approx 1+(a-b)x$$ Then $$A\approx e^{-ax}\frac{1-(1+{(a-b)x}) }{b-a}=-e^{-ax}\frac{{(a-b)x} }{b-a}=x e^{-ax}$$

Answer 3

Consider u=b-a and f(u)=(e^(u-bx) - e^(-bx) ) / u

Limit f(u) when u tends to 0 : using L'hospital : x e^(u x - bx) / 1 = x e^(-bx)

As u tends to 0 , a tends to b , so the limit is x e^(-ax)