4. 0/1 points details my notes find the limit. use lhospitals rule where appropriate. if there is a more…

4. 0/1 points details my notes find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it. lim(x→1) sin(x - 1)/(x³ + 6x - 7)

4. 0/1 points details my notes find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it. lim(x→1) sin(x - 1)/(x³ + 6x - 7)

Answer

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})