1. ## Help in limits

we study limits in calculus so thats why posting here:
if f(x)=Floor[2012*sin(x)/x]
What is Limit x->o[f(x)]?

*Floor function rounds off to least integer. For e.g. Floor[0.99999999]==0.(and not 1)

2. ## Re: Help in limits

Originally Posted by smatik
we study limits in calculus so thats why posting here:
if f(x)=Floor[2012*sin(x)/x]
What is Limit x->o[f(x)]

What do you think the answer is and why?

3. ## Re: Help in limits

Hint :

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x}=1$

4. ## Re: Help in limits

The only tricky part is that sin(x)/x is strictly less than 1. Since floor is discontinuous at every integer, you have to be careful. I hope you can now see the answer and a proof if necessary.

5. ## Re: Help in limits

Now i think its 2011 because sinx/x is slightly less than 1(like 0.9999999999999999999999999999999999...) for x->0 but i've written answer 2012 in my notebook.But i dont remember why i wrote that answer.

6. ## Re: Help in limits

Originally Posted by smatik
Now i think its 2011 because sinx/x is slightly less than 1(like 0.9999999999999999999999999999999999...) for x->0 but i've written answer 2012 in my notebook.But i dont remember why i wrote that answer.