evaluate the limit using lhospitals rule lim x→0 (e^x - 1)/sin(13x)

Explanation:

Step1: Check indeterminate form

As $x\rightarrow0$, $e^{x}-1\rightarrow e^{0}-1 = 0$ and $\sin(13x)\rightarrow\sin(0)=0$. So, it is in $\frac{0}{0}$ form and L'Hopital's rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of $y = e^{x}-1$ is $y'=e^{x}$, and the derivative of $y=\sin(13x)$ using the chain - rule is $y' = 13\cos(13x)$. So, $\lim_{x\rightarrow0}\frac{e^{x}-1}{\sin(13x)}=\lim_{x\rightarrow0}\frac{e^{x}}{13\cos(13x)}$.

Step3: Evaluate the new limit

Substitute $x = 0$ into $\frac{e^{x}}{13\cos(13x)}$. We get $\frac{e^{0}}{13\cos(0)}=\frac{1}{13\times1}=\frac{1}{13}$.

Answer:

$\frac{1}{13}$