Limits

jmsic

New member
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Aug 31, 2009
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Im supposed to evaluate the limit of (sin(pi/6) + delta(x))-1/2 ALL OVER delta x
and incorporate sin(A+B)=sinAcosB+cosAsinB
 
Hello, jmsic!

Your instructions are incomplete . . . I'll take a guess.

I will assume you are given these two theorems:

. . . limθ0sinθθ=1limθ0cosθ1θ=0\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1 \qquad\qquad \lim_{\theta\to0}\,\frac{\cos\theta - 1}{\theta} \:=\:0



We have:   sin(π6+Δx)12  =  sin(π6)cos(Δx)+cos(π6)sin(Δx)12\displaystyle \text{We have: }\;\sin\left(\tfrac{\pi}{6} + \Delta x\right) - \tfrac {1}{2} \;=\;\sin(\tfrac{\pi}{6}) \cos(\Delta x) + \cos(\tfrac{\pi}{6})\sin(\Delta x) - \tfrac{1}{2}

. . . . . . . . . . . . . . . . . . . \(\displaystyle = \;\tfrac{1}{2}\cos(\Dekta x) + \tfrac{\sqrt{3}}{2}\sin(\Delta x) - \tfrac{1}{2}\)


Then:   sin(π6+Δx)12Δx  =  12cos(Δx)+32sin(Δx)12Δx\displaystyle \text{Then: }\;\frac{\sin(\frac{\pi}{6} + \Delta x) - \frac{1}{2}}{\Delta x} \;=\;\frac{\frac{1}{2}\cos(\Delta x) + \frac{\sqrt{3}}{2}\sin(\Delta x) - \frac{1}{2}}{\Delta x}

. . . . . . . . . . =  32sin(Δx)Δx+12cos(Δx)1Δx\displaystyle = \; \frac{\sqrt{3}}{2}\cdot\frac{\sin(\Delta x)}{\Delta x} \:+\: \frac{1}{2}\cdot\frac{\cos(\Delta x) - 1}{\Delta x}


Hence:   limΔx0[32sin(Δx)Δx  +  12cos(Δx)1Δx]\displaystyle \text{Hence: }\;\lim_{\Delta x\to 0}\left[\frac{\sqrt{3}}{2}\cdot\frac{\sin(\Delta x)}{\Delta x} \;+\; \frac{1}{2}\cdot\frac{\cos(\Delta x) - 1}{\Delta x} \right]

. . . . . . . . . . =  32[limΔx0sin(Δx)Δx]This is 1+12[limΔx0cos(Δx)1Δx]This is 0\displaystyle = \;\frac{\sqrt{3}}{2}\cdot\underbrace{\left[\lim_{\Delta x\to0}\frac{\sin(\Delta x)}{\Delta x}\right]}_{\text{This is 1}} + \frac{1}{2}\cdot\underbrace{\left[\lim_{\Delta x\to0}\frac{\cos(\Delta x) - 1}{\Delta x}\right]}_{\text{This is 0}}

. . . . . . . . . . =  321  +  120    =    32\displaystyle = \; \frac{\sqrt{3}}{2}\cdot 1 \;+\;\frac{1}{2}\cdot 0 \;\;=\;\;\boxed{\frac{\sqrt{3}}{2}}
 
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