evaluate the limit using lhôpitals rule lim x...

# Explanation: ## Step1: Check L'Hopital - rule condition When \(x\rightarrow0\), \(\lim_{x\rightarrow0}(e^{x}+3x - 1)=e^{0}+3\times0 - 1=0\) and \(\lim_{x\rightarrow0}(8x)=0\). So, we can apply L'Hopital's rule. ## Step2: Differentiate numerator and denominator The derivative of \(y = e^{x}+3x - 1\) is \(y^\prime=e^{x}+3\), and the derivative of \(y = 8x\) is \(y^\prime = 8\). So, \(\lim_{x\rightarrow0}\frac{e^{x}+3x - 1}{8x}=\lim_{x\rightarrow0}\frac{e^{x}+3}{8}\). ## Step3: Evaluate the new - limit Substitute \(x = 0\) into \(\frac{e^{x}+3}{8}\), we get \(\frac{e^{0}+3}{8}=\frac{1 + 3}{8}\). # Answer: \(\frac{1}{2}\)