Originally Posted by

**jamesrb** $\displaystyle lim_{x\to0}\frac{sin\frac{1}{x}sin^2\frac{x}{2}}{x }$

Now I know I have to get x out of the denominator to get anywhere so I am not dividing by. I tried using the Quotient Law to take the limit of the numerator and denominator separately but quickly realized that I still would be dividing by zero when I go to plug zero in for x. So I guess I could use the Product Law and take two separate limits:

$\displaystyle lim_{x\to0}\frac{sin\frac{1}{x}}{x}lim_{x\to0}sin^ 2\frac{x}{2}$

While I CAN see that because the second part of the equation will equal zero (the sin^2(x/2) part) and thus anything multiplied by zero will equal zero I am at a loss as to further solving the left part of the equation. I once again have x in the denominator. As I mentioned before, I know that using the Quotient Law on the left part will still leave me with x in the denominator and would make it undefined, which doesn't help. I need to get x out of the denominator on the left side, in both the denominator and the (1/x) part of sin. This is where I am having trouble deciding how to go about doing that. Perhaps I shouldn't of even used the Product Law.