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Disclaimer: I am an adult learning Calculus. This is not a student posting his homework assignment. I think this is a great forum!

$$\lim_{h\to0} \frac{\cos(\frac{\pi}{3}+h)-\frac{1}{2}}{h}$$

Do I use the angle addition formula to do this? I did that, and have no idea where to go from there.

$$\lim_{h\to0}\frac{\cos\frac{\pi}{3}\cos(h)-\sin\frac{\pi}{3}\sin(h)-\frac{1}{2}}{h}$$

What now?

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    $\begingroup$ Do you know the definition of a derivative? $\endgroup$
    – Clayton
    Commented Oct 8, 2013 at 16:13
  • $\begingroup$ You are almost done. Note that, $\cos(\pi/3)=1/2$, $\lim_{h\to 0}\frac{\sin(h)}{h}=1$. $\endgroup$ Commented Oct 8, 2013 at 16:28
  • $\begingroup$ So, I factor out a 1/2 and get $\frac{cos(h)-1}{h}$ as one of the limits. Then I'm supposed to know that this limit is 0? (The other one is $\frac{\sqrt(3)}{2}$ $\endgroup$
    – JackOfAll
    Commented Oct 8, 2013 at 18:46

4 Answers 4

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Let $\phi(x) = \cos(\frac{\pi}{3}+x)$. Note that $\phi(0) = \frac{1}{2}$, and $\phi$ is differentiable, with $\phi'(x) = -\sin(\frac{\pi}{3}+x)$.

So, the limit is $\lim_{h \to 0} \frac{\phi(h)-\phi(0)}{h} = \phi'(0) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$.

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  • $\begingroup$ Ok, once you sub the 1/2 for $cos\frac{\pi}{3}$, you just used the limit definition of the derivative! So, you take the derivative of $cos(\frac{\pi}{3})$ and get -sin$\frac{\pi}{3}$ $\endgroup$
    – JackOfAll
    Commented Oct 8, 2013 at 18:40
  • $\begingroup$ Well, we take the derivative of $x \mapsto \cos(\frac{\pi}{3}+x)$ and evaluate it at $x=0$ to get $-\sin \frac{\pi}{3}$. $\endgroup$
    – copper.hat
    Commented Oct 8, 2013 at 21:12
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You might want to use the values of sine and cosine of $\pi/3$ (remember it is 60 degrees), and reduce the problem to evaluating the following two limits: $\lim \frac{1-\cos h}{h}$ and $\lim \frac{\sin h}{h}$. Then note that $1-\cos h$ is a higher order infinitesimal than $h$ so this term drops out. I will leave the second limit to you.

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  • $\begingroup$ How am I supposed to know that $\lim \frac{1-\cos h}{h}=0$ $\endgroup$
    – JackOfAll
    Commented Oct 8, 2013 at 18:50
  • $\begingroup$ Do I just get that limit by viewing the graph around 0 on a graphing calculator? $\endgroup$
    – JackOfAll
    Commented Oct 10, 2013 at 16:32
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$$\frac{d(\cos x)}{dx}=\lim_{h\to0}\frac{\cos(x+h)-\cos x}h=\lim_{u\to0}\frac{\cos(x+2u)-\cos x}{2u}$$

Using $\displaystyle\cos C-\cos D=2\sin\frac{C+D}2\sin\frac{D-C}2$

$$\lim_{u\to0}\frac{\cos(x+2u)-\cos x}{2u}=-\frac12\cdot\lim_{u\to0}\frac{2\sin(x+u)\sin(-u)}{(-u)}=-\lim_{u\to0}\sin(x+u)\cdot\lim_{u\to0}\frac{\sin(-u)}{(-u)}$$

We know $\displaystyle\lim_{x\to0}\frac{\sin x}x=1$

Now in the current context $x=\frac\pi3$ as $\frac12=\cos\frac\pi3,$

Can you take it from here?

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  • $\begingroup$ Can you please point out the mistake here? $\endgroup$ Commented Oct 8, 2013 at 18:44
  • $\begingroup$ Looks good to me. $\endgroup$
    – copper.hat
    Commented Oct 8, 2013 at 21:10
  • $\begingroup$ I just downvoted it since it seemed more complicated than necessary. I assume it's correct. Thanks! $\endgroup$
    – JackOfAll
    Commented Oct 10, 2013 at 16:32
  • $\begingroup$ @JackOfAll, does it still look complicated? Find here :mash.dept.shef.ac.uk/Resources/sincosfirstprinciples.pdf $\endgroup$ Commented Oct 10, 2013 at 17:09
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In this case, we can also use L'Hospital Rule because $\lim_{h \to 0} h =0$ and $\lim_{h \to 0} (cos(\pi/3+h)-1/2) =(cos(\pi/3+0)-1/2)=0$. $$\lim_{h \to 0}\frac{cos(\pi/3+h)-1/2}{h} =\frac{\lim_{h \to 0}(\frac{d}{dh}(cos(\pi/3+h)-1/2))}{\lim_{h \to 0}(\frac{d}{dh}h)}=\frac{\lim_{h \to 0}(-sin(\pi/3+h))}{\lim_{h \to 0}1}=\frac{-sin(\pi/3+0)}{1}=-\sqrt{3}/2.$$

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