Question

Solve $\underset{x\to 1}{lim}\frac{\sin (x-1)}{(x^2-1)}$

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $\frac{1}{5}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}$

$\Rightarrow \underset{x\to 1}{lim}\frac{\sin (x-1)}{x^2-1}$

$\Rightarrow \underset{x\to 1}{lim}\frac{\sin (x-1)}{x-1}.\underset{x\to 0}{lim}\frac{1}{x+1}$

$\Rightarrow 1.\underset{x\to 0}{lim}\frac{1}{x+1}=\frac{1}{2}$