Quote (Arturo @ Jan 13 2010 04:00pm)
Find the limit.
Codelim sin[(π/6) + Δx] - (1/2)
-------------------------
Δx -> 0 Δx
[Hint: cos (θ + Φ) = sin θ cos Φ - cos θ sin Φ]
[The dashes are a fraction bar. In case you couldn't figure that out.]
I need this problem solved with you showing the work. I have tried it and can't figure out what to do after using the hint. I know what the correct answer is, but do not know how to get it.
I will donate you some fg if you do this problem correctly and showing the work. I will know if your final answer is right or wrong, so please don't just make stuff up.
Thanks.
Code
Since sin(Pi/6)-x)-1/2 -> 0 as x->0, you need to use L'hospitals rule :
lim sin[(π/6) + Δx] - (1/2) lim cos[(π/6) + Δx])
------------------------- = --------------------- = cos(π/6) =sqrt(3)/2
Δx -> 0 Δx Δx -> 0 1
Oh i guess you guys don't know about L'hopitals rule yet lol.
Quote (xbeastpwnerx @ Jan 13 2010 04:44pm)
got it...
knowing sinx/x = 0 (this should be given, I can prove this if need be)
sin(pi/6 + x) - 1/2 = sin(pi/6)cos(x) + cos(pi/6)sin(x) - 1/2 = 1/2cos(0) + sqrt(3)/2 * sin(x) - 1/2
____ x__________________________x______________________________x
simplifying the above I get:
sqrt(3)/2 * sin(x) lim x--> 0 = sqrt(3)/2
______ x
sin(x)/x=1 when x->0
This post was edited by Taxidermy on Jan 13 2010 03:53pm