### Related Formula The fundamental trigonometric limit is given by: lim_x to 0 fracsin xx = 1$\lim_{x \to 0} \frac{\sin x}{x} = 1$ ### Core Logic Apply the limit inside each element of the matrix determinant : lim_x to 0 f(x) = beginvmatrix a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 endvmatrix$\lim_{x \to 0} f(x) = \begin{\vmatrix} a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 \end{\vmatrix}$ ### Step 1: Simplify the Determinant via Row Operations Perform rows reductions R_2 to R_2 - R_1$R_2 \to R_2 - R_1$ and R_3 to R_3 - R_1$R_3 \to R_3 - R_1$ to create zeros: lim_x to 0 f(x) = beginvmatrix a+1 & 1 & b \\ -1 & 1 & 0 \\ -1 & 0 & 1 endvmatrix$\lim_{x \to 0} f(x) = \begin{\vmatrix} a+1 & 1 & b \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{\vmatrix}$ Expand across the first row : = (a+1)[1(1) - 0] - 1[-1(1) - 0] + b[0 - (-1)]$= (a+1)[1(1) - 0] - 1[-1(1) - 0] + b[0 - (-1)]$ = (a+1)(1) + 1 + b = a + b + 2$= (a+1)(1) + 1 + b = a + b + 2$ [cite: 3999, 4000] ### Step 2: Match Coefficients Equate this outcome with the target parameter template lambda + mu a + nu b$\lambda + \mu a + \nu b$ [cite: 3377, 4000]: lambda = 2, quad mu = 1, quad nu = 1$\lambda = 2, \quad \mu = 1, \quad \nu = 1$ Calculate (lambda + mu + nu)^2$(\lambda + \mu + \nu)^2$ [cite: 3378, 4001]: (2 + 1 + 1)^2 = 4^2 = 16$(2 + 1 + 1)^2 = 4^2 = 16$ ### Pattern Recognition Standard row manipulations on identity-shifted arrays quickly eliminate complex parameter symbols, reducing determinant calculations into straightforward polynomial expansions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Limits and Derivatives