# Thread: Having trouble with trig based limits

1. ## Having trouble with trig based limits

As h approaches 0,

lim [cos((pi/2)+h) - cos(pi/2)] / h

Are there special trig-based formulas I need to know to solve questions like this?

2. ## Re: Having trouble with trig based limits

Originally Posted by TWN
As h approaches 0,

lim [cos((pi/2)+h) - cos(pi/2)] / h

Are there special trig-based formulas I need to know to solve questions like this?
Two special limits you should know are

$\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$ and $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x}=0$.

In this case, we have

$\displaystyle \lim_{h\to0}\frac{\cos\left(\frac{\pi}2+h\right)-\cos\frac{\pi}2}h$

$\displaystyle =\lim_{h\to0}\frac{\cos\frac\pi2\cos h-\sin\frac\pi2\sin h-0}h$ (sum and difference identity)

$\displaystyle =\lim_{h\to0}\frac{0\cdot\cos h-1\cdot\sin h}h$

$\displaystyle =\lim_{h\to0}\frac{-\sin h}h$

$\displaystyle =-\lim_{h\to0}\frac{\sin h}h$

$\displaystyle =-1$

3. ## Re: Having trouble with trig based limits

Ah, I see. I was unaware of the sum and difference identities. Thank you!

By the way, how do you write out the problems like that? It's so much easier to read than mine is!

4. ## Re: Having trouble with trig based limits

Originally Posted by TWN
Ah, I see. I was unaware of the sum and difference identities. Thank you!

By the way, how do you write out the problems like that? It's so much easier to read than mine is!
You can write in $\displaystyle \LaTeX$, just enclose your code between $$and$$ tags. For help with LaTeX, visit the LaTeX forum. There's a tutorial there.

5. ## Re: Having trouble with trig based limits

Originally Posted by TWN
As h approaches 0,

lim [cos((pi/2)+h) - cos(pi/2)] / h

Are there special trig-based formulas I need to know to solve questions like this?
note the format of the limit ...

$\displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

... this is the definition of $\displaystyle f'(x)$

in this case, $\displaystyle f(x) = \cos{x}$ and the limit is $\displaystyle f'\left(\frac{\pi}{2}\right) = -\sin \left(\frac{\pi}{2}\right) = -1$

6. ## Re: Having trouble with trig based limits

That makes sense! Thank you both!