Evaluate lim(x→0) (e^bx - e^ax)/x, 0 < a < b

Question

Evaluate \(\lim\limits_{\text x \to0}\left(\cfrac{e^{b\text x}-e^{a\text x}}{\text x}\right), \) 0 < a < b

lim(x→0) (e^bx - e^ax)/x, 0 < a < b

Answer 1

To evaluate : \(\lim\limits_{\text x \to0}\left(\cfrac{e^{b\text x}-e^{a\text x}}{\text x}\right), \) 0 < a < b

lim(x→0) (e^bx - e^ax)/x, 0 < a < b

Formula used: L'Hospital's rule

Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

Thus, the value of \(\lim\limits_{\text x \to0}\cfrac{e^{b\text x}-e^{a\text x}}{\text x}\) is b - a.