Question

Evaluate the following limits.
$\underset{x\to 0}{lim}\frac{e^{3x}-e^{2x}}{x}$.

Answer 1

We know $\underset{x\to 0}{lim}\frac{e^x-1}{x}=1$

$\underset{x\to 0}{lim}\frac{e^{3x}-e^{2x}}{x}=\underset{x\to 0}{lim}\frac{e^{3x}-1+1-e^{2x}}{x}$

$=\underset{x\to 0}{lim}\frac{e^{3x}-1}{x}-\underset{x\to 0}{lim}\frac{e^{2x}-1}{x}$

$=\underset{x\to 0}{lim}3.\frac{e^{3x}-1}{3x}-\underset{x\to 0}{lim}2.\frac{e^{2x}-1}{2x}$

$=3.1-2.1=3.2$

$=1$

Hence $\underset{x\to 0}{lim}\frac{e^{3x}-e^{2x}}{x}=1$