evaluate the limit using lhopitals rule lim x...

# Explanation: ## Step1: Check indeterminate form As $x\rightarrow\infty$, we have $\frac{14x^{3}}{e^{4x}}\rightarrow\frac{\infty}{\infty}$, so L'Hopital's rule can be applied. ## Step2: Differentiate numerator and denominator Differentiate $14x^{3}$ to get $42x^{2}$ and $e^{4x}$ to get $4e^{4x}$. So the limit becomes $\lim_{x\rightarrow\infty}\frac{42x^{2}}{4e^{4x}}=\lim_{x\rightarrow\infty}\frac{21x^{2}}{2e^{4x}}$. ## Step3: Apply L'Hopital's rule again Differentiate $21x^{2}$ to get $42x$ and $2e^{4x}$ to get $8e^{4x}$. So the limit becomes $\lim_{x\rightarrow\infty}\frac{42x}{8e^{4x}}=\lim_{x\rightarrow\infty}\frac{21x}{4e^{4x}}$. ## Step4: Apply L'Hopital's rule one - more time Differentiate $21x$ to get $21$ and $4e^{4x}$ to get $16e^{4x}$. So the limit becomes $\lim_{x\rightarrow\infty}\frac{21}{16e^{4x}}$. ## Step5: Evaluate the limit As $x\rightarrow\infty$, $e^{4x}\rightarrow\infty$, so $\lim_{x\rightarrow\infty}\frac{21}{16e^{4x}} = 0$. # Answer: $0$