Calculus I problem. Evaluating a limit.

[jamesrb]

\[
lim_{x\to0}\frac{sin\frac{1}{x}sin^2\frac{x}{2}}{x}
\]


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:

\[
lim_{x\to0}\frac{sin\frac{1}{x}}{x}lim_{x\to0}sin^2\frac{x}{2}
\]

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.

 

[tkhunny]

I still would be dividing by zero when I go to plug zero in for x.

Well, there's your problem, right there. Please forget the time-honored tradition of "plugging in". This is a LIMIT. You are APPROACHING. If you arive, you are no longer approaching. Seriously, stop thinking that way.

\(\displaystyle \frac{\sin(\frac{x}{2})}{x} = \frac{\frac{1}{2}\cdot \sin(\frac{x}{2})}{\frac{x}{2}}\)

That's a good piece. Think about it as you ponder \(\displaystyle \frac{\left(\sin(\frac{x}{2})\right)^{2}}{x} = \frac{x\cdot \left(\sin(\frac{x}{2})\right)^{2}}{x^{2}}\)

 

[HallsofIvy]

\(\displaystyle \frac{sin(1/x) sin^2(x/2)}{x}= \frac{1}{x} sin(1/x)sin(x/2)sin(x/2)\)
\(\displaystyle = \frac{1}{x}\frac{sin(1/x)}{\frac{1}{x}}\)\(\displaystyle \left(\frac{1}{x}\right)\)\(\displaystyle \frac{sin(x/2)}{\frac{x}{2}}\left(\frac{x}{2}\right)\)\(\displaystyle \frac{sin(x/2)}{\frac{x}{2}}\left(\frac{x}{2}\right)\)
\(\displaystyle =\frac{1}{x}\frac{1}{x}\frac{x}{2}\frac{x}{2}\)\(\displaystyle \frac{sin(1/x)}{\frac{1}{x}}\)\(\displaystyle \frac{sin(x/2)}{\frac{x}{2}}\)\(\displaystyle \frac{sin(x/2)}{\frac{x}{2}}\)
\(\displaystyle = \frac{1}{4}\frac{sin(1/x)}{\frac{1}{x}}\)\(\displaystyle \frac{sin(x/2)}{\frac{x}{2}}\)\(\displaystyle \frac{sin(x/2)}{\frac{x}{2}}\)

Now take the limit of each of those. You cannot just put 0 in for x in those fractions- you have to KNOW the limit of \(\displaystyle \frac{sin(u)}{u}\) as u goes to 0 or as u goes to infinity.