Limits

[jmsic]

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

 

[soroban]

Hello, jmsic!

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

I will assume you are given these two theorems:

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

\(\displaystyle \text{Evaluate: }\:\lim_{\Delta x\to0}\,\frac{\sin(\frac{\pi}{6} + \Delta x) - \frac{1}{2}}{\Delta x}\)

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

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


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

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


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

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

. . . . . . . . . . \(\displaystyle = \; \frac{\sqrt{3}}{2}\cdot 1 \;+\;\frac{1}{2}\cdot 0 \;\;=\;\;\boxed{\frac{\sqrt{3}}{2}}\)