Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 151: 50

$$\frac{1}{3}$$

Given $$\lim _{x\to 1}\frac{\sin \left(x-1\right)}{x^2+x-2}$$ Since \begin{align*} \lim _{x\to 1}\frac{\sin \left(x-1\right)}{x^2+x-2}&=\lim _{x\to 1}\frac{\sin \left(x-1\right)}{(x-1)(x+2)}\\ &=\lim _{x\to 1}\frac{1}{x+2}\lim _{x\to 1}\frac{\sin \left(x-1\right)}{(x-1)}\\ &=\lim _{x\to 1}\frac{1}{x+2}\lim _{x-1\to 0}\frac{\sin \left(x-1\right)}{(x-1)}\\ &=\frac{1}{3}(1)==\frac{1}{3} \end{align*}