4. 0/1 points details my notes find the limit...

# Explanation: ## Step1: Check form of limit Substitute \(x = 1\) into \(\frac{\sin(x - 1)}{x^{3}+6x - 7}\). We get \(\frac{\sin(1 - 1)}{1^{3}+6\times1 - 7}=\frac{\sin(0)}{1 + 6-7}=\frac{0}{0}\), an indeterminate - form. ## Step2: Apply L'Hopital's Rule Differentiate the numerator and denominator. The derivative of \(y=\sin(x - 1)\) using the chain - rule is \(y^\prime=\cos(x - 1)\), and the derivative of \(y=x^{3}+6x - 7\) is \(y^\prime = 3x^{2}+6\). So, \(\lim_{x\rightarrow1}\frac{\sin(x - 1)}{x^{3}+6x - 7}=\lim_{x\rightarrow1}\frac{\cos(x - 1)}{3x^{2}+6}\). ## Step3: Evaluate the new limit Substitute \(x = 1\) into \(\frac{\cos(x - 1)}{3x^{2}+6}\). We have \(\frac{\cos(1 - 1)}{3\times1^{2}+6}=\frac{\cos(0)}{3 + 6}=\frac{1}{9}\). # Answer: \(\frac{1}{9}\)