evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim x→0 (e^3x…

Explanation:

Step1: Check indeterminate form

When $x\rightarrow0$, we have $\lim_{x\rightarrow0}(e^{3x}-1)=e^{0}-1 = 0$ and $\lim_{x\rightarrow0}(2x^{2}+9x)=0$. So it is in the $\frac{0}{0}$ - indeterminate form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $y = e^{3x}-1$ is $y^\prime=3e^{3x}$ (by the chain - rule, if $y = e^{u}$ and $u = 3x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=e^{u}\cdot3 = 3e^{3x}$), and the derivative of $y = 2x^{2}+9x$ is $y^\prime=4x + 9$. So $\lim_{x\rightarrow0}\frac{e^{3x}-1}{2x^{2}+9x}=\lim_{x\rightarrow0}\frac{3e^{3x}}{4x + 9}$.

Answer:

$\frac{3e^{0}}{4\times0 + 9}=\frac{3}{9}=\frac{1}{3}$