Find the limit. Use I'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. \[ \lim _{x \rightarrow 1} \frac{\sin (x-1)}{x^{3}+6 x-7} \] \( \square \)
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To find the limit \[ \lim _{x \rightarrow 1} \frac{\sin (x-1)}{x^{3}+6 x-7, \] we first evaluate the function at \(x = 1\): The numerator becomes \(\sin(1-1) = \sin(0) = 0\). The denominator becomes \((1)^3 + 6(1) - 7 = 1 + 6 - 7 = 0\). Thus, we have an indeterminate form \(\frac{0}{0}\), so we can apply L'Hôpital's Rule. We take the derivative of the numerator and the denominator: 1. Derivative of the numerator \(\sin(x-1)\) is \(\cos(x-1) \cdot 1 = \cos(x-1)\). 2. Derivative of the denominator \(x^3 + 6x - 7\) is \(3x^2 + 6\). Now, we can rewrite the limit: \[ \lim_{x \rightarrow 1} \frac{\cos(x-1)}{3x^2 + 6}. \] Substituting \(x = 1\): The numerator becomes \(\cos(1-1) = \cos(0) = 1\). The denominator becomes \(3(1)^2 + 6 = 3 + 6 = 9\). So, we have: \[ \frac{1}{9}. \] Thus, the limit is \[ \boxed{\frac{1}{9}}. \]
