10. - / 1 points find the limit. $lim_{x\rightarrow1}\frac{sin(x - 1)}{x^{2}+x - 2}$

Explanation:

Step1: Factor the denominator

Factor $x^{2}+x - 2=(x + 2)(x - 1)$. So the limit becomes $\lim_{x\rightarrow1}\frac{\sin(x - 1)}{(x + 2)(x - 1)}$.

Step2: Use the limit - property $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$

Let $u=x - 1$. As $x\rightarrow1$, $u\rightarrow0$. Then $\lim_{x\rightarrow1}\frac{\sin(x - 1)}{(x + 2)(x - 1)}=\lim_{u\rightarrow0}\frac{\sin u}{(u + 3)u}=\lim_{u\rightarrow0}\frac{\sin u}{u}\cdot\frac{1}{u + 3}$.

Step3: Evaluate the limit

Since $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$ and $\lim_{u\rightarrow0}\frac{1}{u + 3}=\frac{1}{0 + 3}=\frac{1}{3}$, then $\lim_{u\rightarrow0}\frac{\sin u}{u}\cdot\frac{1}{u + 3}=1\times\frac{1}{3}=\frac{1}{3}$.

Answer:

$\frac{1}{3}$