# Derivatives of trig functions

• Nov 12th 2010, 05:28 AM
brennon
Derivatives of trig functions
Having a ball teaching myself calculus for DSP...

I'm working with derivatives of trig functions. I'm trying to solve the following limit:

$\displaystyle \lim_{x\rightarrow 0} \frac{x \csc 2x}{\cos 5x}$

My first thought was to just set x to 0, since the equation would still remain valid. This is obviously wrong, when checking the answer in the back of the book, which gives this limit as 1/2. This question immediately follows a proof regarding the limit of sin(x)/x as x approaches zero always being 1, so I thought it might be helpful to look for ways to reduce this to something along those lines, but I'm just not seeing it. Can someone point me in the right direction?

Many thanks,
Brennon
• Nov 12th 2010, 05:34 AM
Plato
You should realize that:
$\displaystyle \displaystyle x\cdot\csc(2x)=\dfrac{2x}{2\cdot \sin(2x)}$
that limit is $\displaystyle \frac{1}{2}$
• Nov 12th 2010, 06:14 AM
brennon
Well, I had gotten far enough to realise that

$\displaystyle x\csc{2x} = x\dfrac{1}{\sin 2x}$

which is essentially what you've posted. What I don't see, however, is how you take the limit of this as x approaches zero to be $\displaystyle \frac{1}{2}$. I don't see, unless my following logic here is correct (but it seems fishy to me):

$\displaystyle \lim_{x\rightarrow 0}\dfrac{2x}{2\sin (2x)} = \lim_{x\rightarrow 0}\dfrac{1}{2}\cdot \dfrac{2x}{\sin (2x)} = \dfrac{1}{2}\cdot\lim_{x\rightarrow 0}\dfrac{2x}{\sin (2x)}$

Furthermore, since

$\displaystyle \lim_{x\rightarrow 0}\dfrac{\sin x}{x} = 1$

$\displaystyle \lim_{x\rightarrow 0}\dfrac{x}{\sin x} = 1$

Thus,

$\displaystyle \dfrac{1}{2}\cdot\lim_{x\rightarrow 0}\dfrac{2x}{\sin (2x)} = \dfrac{1}{2}\cdot {1} = \dfrac{1}{2}$

Am I on the right track?

Thanks again,
Brennon
• Nov 14th 2010, 06:48 AM
brennon
Knowing I'm doing the right thing is just as helpful as knowing I've done something stupid... ;) Anyone?
• Nov 14th 2010, 08:00 AM
drumist
Quote:

Originally Posted by brennon
My first thought was to just set x to 0, since the equation would still remain valid.

It actually doesn't remain valid. The reason is that $\displaystyle \csc 2x$ goes toward $\displaystyle \pm\infty$ as $\displaystyle x\to0$. So you're left with a term $\displaystyle x$ that goes to 0 and a term $\displaystyle \csc 2x$ that goes toward $\displaystyle \infty$ multiplied together... in other words, this is the $\displaystyle 0 \cdot \infty$ indeterminate form.

That's why direct substitution does not work.

Quote:

Originally Posted by brennon
Am I on the right track?

Yes, you did it correctly.