Evaluating Limit, h->0, [sin((pi/4) + h) - sqrt(2)/2] / h

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jmorris1

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Dec 7, 2008
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The problem I have is:
lim (sin((pi/4) + h) - sqrt(2)/2)/h
h-->0

I tried to simplify the numerator. I made it sin[(pi+4h)/4] - sqrt (2)/2. I thought the next step would be to rationalize the numerator. Don't know if my algebra was messed up or if I am doing it wrong. Regardless, thanks for any help I can get!
 
Re: Evaluating Limits

jmorris1 said:
The problem I have is:
lim (sin((pi/4) + h) - sqrt(2)/2)/h
h-->0

I tried to simplify the numerator. I made it sin[(pi+4h)/4] - sqrt (2)/2. I thought the next step would be to rationalize the numerator. Don't know if my algebra was messed up or if I am doing it wrong. Regardless, thanks for any help I can get!
Click to expand...

Use:

\(\displaystyle \sin (A+B) \, = \, \sin A \cdot \cos B \, + \, \sin B \cdot \cos A\)
 
Re: Evaluating Limits

jmorris1 said:
why do you use that? I am a little bit confused.
Click to expand...

take a look at your numerator carefully - you have -

\(\displaystyle \sin (\frac{\pi}{4} \, + \, h)\)

you can expand that according to the formula given before.
 
Re: Evaluating Limits

okay, I vaguely remember doing something similar to that before. I did that and came up with sqrt (2)/2 * [cos(h) +sin(h) -1]/h. from there I can't see where to go with the h in the denominator. I see no way of getting rid of it, and leaving it results in a limit that does not exist. So where does that go?
 
hint:

\(\displaystyle \text Lim_{h->0} \frac{\sin(h)}{h} \, = \, 1\)
 
jmorris1 said:
The problem I have is:
lim (sin((pi/4) + h) - sqrt(2)/2)/h
h-->0

I tried to simplify the numerator. I made it sin[(pi+4h)/4] - sqrt (2)/2. I thought the next step would be to rationalize the numerator. Don't know if my algebra was messed up or if I am doing it wrong. Regardless, thanks for any help I can get!
Click to expand...
.....................................
Isn't sin(pi/4) = sqrt(2)/2?

Let's call pi/4 = x for now.

Then your limit is:

sin(x + h) - sin(x)
lim ---------------------
h->0 h

Does it look familiar? Did you read the posts for your previous question?
 
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