Question

$\underset{x\to 0}{\lim }\frac{e^{bx}-e^{ax}}{x}\mathrm{where}0<a<b$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{e^{bx}-e^{ax}}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\left(\frac{e^{bx}-1}{x}\right)-\left(\frac{e^{ax}-1}{x}\right)\right] \\ =\underset{x\to 0}{\lim }\left[b\left(\frac{e^{bx}-1}{bx}\right)-a\times \left(\frac{e^{ax}-1}{ax}\right)\right] \\ =b\times 1-a\times 1 \\ =b-a \end{gathered}$$