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$\lim_{x\to 1} \frac{\sin (x-1)}{x-1}$

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.

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Maybe try using sin series expansion? Or L'Hospital – Stefan Oct 8 '12 at 19:23
Try substitution $t=x-1$. – M. Strochyk Oct 8 '12 at 19:25
up vote 4 down vote accepted

You could simply write down pretty much just what you’ve written here: it shows that you understand why the limit is $1$. However, a nice way is to make a substitution $y=x-1$; then clearly

$$\lim_{x\to 1}\frac{\sin(x-1)}{x-1}=\lim_{y+1\to 1}\frac{\sin y}y=\lim_{y\to 0}\frac{\sin y}y=1\;.$$

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