limit problem

jimmym2212

New member
Joined
Oct 23, 2005
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5
ok I'm stuck.

limit as delta x approaches 0 for ( sin(pie/6 + delta X) - 1/2 ) / delta x

I know the limit is sqroot of 3/2 because its in the back of the book but I'm having trouble doing the work. This is my last resort. It's due tomorrow and counts for 25 points on a quarter exam. Any help is appreciated.
 
Have you tried expanding the sine functon with the sum formula?

sin(x+y) = sin(x)cos(y) + sin(y)cos(x)

That might help.
 
How one does this depends on the tools at hand.
Do you know that Lim<SUB>h->0</SUB>[sin(h)/h]=1?
Do you know that Lim<SUB>h->0</SUB>[(cos(h)−h)/h]=0?
If you have these, expand sin([pi]/6 + h)=sin([pi]/6)cos(h)+sin(h)cos([pi]/6).
You know sin([pi]/6)=1/2 and cos([pi]/6)=(√3)/2.
 
tkhunny and pka,

I've expanded it out with the sum formula. Here's what i have so far.

( 2sin(pi/6)cos(delta x) + 2cos(pi/6)sin(delta x) -1 ) / 2 delta x

I can't figure the rest out.

Thanks
 
2sin(pi/6)cos(h) + 2cos(pi/6)sin(h) -1 ) / 2h =[cos(h)−1]/[2h] + cos(pi/6)[sin(h)/h]
 
pka said:
2sin(pi/6)cos(h) + 2cos(pi/6)sin(h) -1 ) / 2h =[cos(h)−1]/[2h] + cos(pi/6)[sin(h)/h]


ok so it would be [cos(h)/1] / [2h] + cos(pi/6)[sin(h)/h]

which is
1/ 2h + sqroot 3/2 [0/0]

or 1/0 + sqroot 3/2 [0/0]

any closer? Sorry if I'm not too good at this but the help is appreciated
 
It appears that you are going to just miss this one. Sorry.
 
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