Answer
3
Work Step by Step
We have to compute $\lim\limits_{x \to 0}\dfrac{e^{3x}-1}{x}$.
We use a calculator to evaluate the function $f(x)=\dfrac{e^{3x}-1}{x}$ for values of $x$ very close to 0.
Let $x_1=0.00001,x_2=0.000001, x_3=0.0000001$.
$f(x_1)=f(0.00001)=\dfrac{e^{3(0.00001)}-1}{0.00001}=3.00004500045$
$f(x_1)=f(0.000001)=\dfrac{e^{3(0.000001)}-1}{0.000001}=3.00000449993$
$f(x_1)=f(0.0000001)=\dfrac{e^{3(0.0000001)}-1}{0.0000001}=3.00000045028$
Therefore we have:
$\lim\limits_{x \to 0}\dfrac{e^{3x}-1}{x}=3$
