Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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.

share|improve this question
    
Maybe try using sin series expansion? Or L'Hospital –  Stefan Oct 8 '12 at 19:23
3  
Try substitution $t=x-1$. –  M. Strochyk Oct 8 '12 at 19:25
add comment

1 Answer

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.