# siny/y ? ...

• Apr 10th 2010, 06:53 AM
matlabnoob
siny/y ? ...
can anyone explain why...

lim (siny/y) = 1 , as y --> 0

?... surely siny/y = 0... as y--->0

because sin0/0 = 0 ??

thanks!
• Apr 10th 2010, 07:56 AM
Because for small x, $\sin x \approx x$...

Here's a proof Khan Academy: Proof: lim (sin x)/x which will explain it better than I can.
• Apr 10th 2010, 08:04 AM
drumist
While you may believe the function is equal to zero because the numerator of the faction is zero, consider that you are not allowed to divide any number by zero, which is what happens if you attempt to evaluate $\frac{\sin x}{x}$ at x=0.

$\frac{0}{0}$ is what is called an indeterminate form. The numerator demands that the value be zero, but the denominator demands that the expression is tending toward infinity. But both can't be true at the same time, though, so we have to treat it differently. In any case, what we know is the function is undefined at x=0, so evaluating it at that point does not give us any information at all.

Remember that a limit is determined by what happens AROUND a point, and completely ignores what actually happens AT the point. Take a look at the graph on page 2 of this pdf: http://pup.princeton.edu/books/maor/chapter_10.pdf

The rest of the article looks more advanced than what you are looking for, so you might not want to read too much of it, but the graph at least is helpful to see what is going on.
• Apr 10th 2010, 08:12 AM
mathemagister
As aforementioned, 0/0 is indeterminate (see Indeterminate -- from Wolfram MathWorld).

If you are familiar with L'Hospital's Rule (if not, see L'Hospital's Rule -- from Wolfram MathWorld), you will notice that

$\lim_{y\rightarrow 0} \frac{siny}{y} = \lim_{y\rightarrow 0} \frac{cosy}{1} = \frac{1}{1} = 1$

Hope that helps

Mathemagister
• Apr 10th 2010, 04:03 PM
matlabnoob
thank you everyone!! i got it(Wink)