evaluate the limit using lhospitals rule\nlim...

# Explanation: ## Step1: Check form of limit As $x\rightarrow0$, $\lim_{x\rightarrow0}\frac{e^{x}-1}{\sin(13x)}=\frac{e^{0}-1}{\sin(0)}=\frac{1 - 1}{0}=\frac{0}{0}$, so 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 $(u = 13x,y=\sin(u),\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx})$ 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: $\frac{1}{13}$