evaluate the limit using lhospitals rule lim ...

# Explanation: ## Step1: Check indeterminate form When $x\rightarrow0$, $\sin(15x)\rightarrow0$ and $\tan(6x)\rightarrow0$. So, it's a $\frac{0}{0}$ - form, and L'Hopital's rule can be applied. ## Step2: Differentiate numerator and denominator The derivative of $y = \sin(15x)$ using the chain - rule is $y^\prime=15\cos(15x)$. The derivative of $y=\tan(6x)$ using the chain - rule is $y^\prime = 6\sec^{2}(6x)$. So, $\lim_{x\rightarrow0}\frac{\sin(15x)}{\tan(6x)}=\lim_{x\rightarrow0}\frac{15\cos(15x)}{6\sec^{2}(6x)}$. ## Step3: Evaluate the new limit When $x\rightarrow0$, $\cos(15x)\rightarrow1$ and $\sec^{2}(6x)\rightarrow1$. Then $\lim_{x\rightarrow0}\frac{15\cos(15x)}{6\sec^{2}(6x)}=\frac{15\times1}{6\times1}=\frac{5}{2}$. # Answer: $\frac{5}{2}$