# Having trouble with trig based limits

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• May 14th 2012, 05:26 PM
TWN
Having trouble with trig based limits
As h approaches 0,

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

The answer is -1. Can someone please explain how to arrive at this answer?

Are there special trig-based formulas I need to know to solve questions like this?
• May 14th 2012, 05:47 PM
Reckoner
Re: Having trouble with trig based limits
Quote:

Originally Posted by TWN
As h approaches 0,

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

The answer is -1. Can someone please explain how to arrive at this answer?

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)

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

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

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

$\displaystyle =-1$
• May 14th 2012, 06:32 PM
TWN
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!
• May 14th 2012, 06:37 PM
Reckoner
Re: Having trouble with trig based limits
Quote:

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.
• May 15th 2012, 03:06 AM
skeeter
Re: Having trouble with trig based limits
Quote:

Originally Posted by TWN
As h approaches 0,

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

The answer is -1. Can someone please explain how to arrive at this answer?

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 $\displaystyle f'\left(\frac{\pi}{2}\right) = -\sin \left(\frac{\pi}{2}\right) = -1$
• May 15th 2012, 02:16 PM
TWN
Re: Having trouble with trig based limits
That makes sense! Thank you both!