Question

Show that the limit of [e^(-ax)-e^(-bx)]/x=b-a (while x->0)

Answer 1

Let f(x) = e^(-ax)-e^(-bx) and g(x) = x. The limit as x tends to 0 of both f and g is 0 and so we cannot deduce the limit directly. L'Hopital's Rule states that in such cases the limit is given by
\[\lim_{x \rightarrow 0}\left\{ f \prime(x)/g \prime(x) \right\} \]Since
\[f \prime(x)=-ae ^{-ax}+be ^{-ay} \]and
\[g \prime(x) = 1\]then the limit is b-a as stated