$\lim_{x\to 1} \frac{\sin (x-1)}{x-1}$
I know the answer equals $1$ because $\lim_{x\to 0} \frac{\sin (x)}{x} = 1$ and in the following question $x-1$ gets arbitrary close to 0 so the same thing is happening. What I need is some steps to basically show that the question was not solved by a calculator.
I tried to use $\sin(A-B) = \sin A \mathrm{cos}B - \sin B \cos A $ but I had a $\frac {0}0$ which is obviously wrong. Any help/tip would be great.