Compute $\cos'(0)$

I know that $\cos'(0)$ is $0$, but my work follows:

\begin{align}\cos'(0) &= \lim_{\Delta x\to 0}\frac{f(0+\Delta x) - f(0)}{\Delta x}\\ &= \frac{\cos(0+\Delta x) - \cos(0)}{\Delta x}\\ &= \frac{\cos(0) - \cos(0)}{0} = \frac{0}{0}\end{align}

Where have I gone wrong? What can I do to show that it equals 0?

(Note: No derivative rules are allowed in my calculus class yet, just difference quotients)

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Thanks for clearing that up, I copied it down incorrectly. Even with 0 instead of $\Delta x$, I still end up with $\frac{0}{0}$ –  BKaylor Sep 11 '11 at 19:11
What esle did you do wrong? You "dropped" the lim in front, with no reason. Then you set $\Delta x$ equal to $0$ with no reason. But you will still need to know some non-trivial property of trig functions. –  GEdgar Sep 11 '11 at 19:12

You can't just substitute in 0 for the limit - that loses all important information and leaves you with an indeterminate form. You must be wittier - usually, with $\cos$ and $\sin$ and their derivatives (even with difference quotients), one ends up using the following inequalities.

$$\cos A - \cos B = -2 \sin {\frac{1}{2} (A + B)} \cos {\frac{1}{2} (A + B)}$$

$$\cos(x) = \sin(\frac{\pi}{2} - x)$$

Or you could use the difference of $\sin$ angles too, depending on how you tackle the proof. The key is to not get rid of the $\Delta x$ too early - evaluate the limit only when you come across a form that you know how to evaluate.

I think that gives you a next step, right?

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@BKaylor One may also apply the double-angle formulas: $\cos(2\theta) = 1-2 \sin^2 \theta$. (Of course, this can also be derived from the formulas given in this answer.) –  Srivatsan Sep 11 '11 at 19:24
The main things I'm supposed to use are algebraic limit manipulations ( like lim f(x) + g(x) = lim f(x) + lim g(x) ) And "special" trig limits that we've worked out in class. We've been told that we don't need to prove the two limits $\lim_{x\to 0}\frac{\sin(x)}{x}$ and $\lim_{x\to 0}\frac{1- \cos(x)}{x}$, just to assume they equal 1 and 0 respectively. I think the proof $\lim_{\Delta x\to 0}\frac{[\lim_{\Delta x\to 0} \cos(0 + \Delta x)] - \cos(\Delta x)}{\Delta x} = 0$ Should work... thanks for pointing me in the right direction! –  BKaylor Sep 11 '11 at 19:40
@Thijs All sum-angle and sum-difference formulas, like the ones in the answer, can be derived geometrically (sometimes this is done using vectors in the plane). One can of course get the formulas for $\sin 2x$, $\cos 2x$ and so on as special cases. See here. –  Srivatsan Sep 11 '11 at 19:42

Because it is easier to type, I will write $h$ instead of $\Delta x$. We want to find $$\lim_{h \to 0} \frac{\cos h -1}{h}.$$ Multiply "top" and "bottom" by $\cos h+1$. We want $$\lim_{h \to 0} \frac{(\cos h -1)(\cos h+1)}{h(\cos h+1)}.$$ On top we now have $-\cos^2 h +1$, that is, $-\sin^2 h$. So we want $$\lim_{h \to 0} \frac{-\sin^2 h}{h(\cos h+1)}\qquad\text{that is,}\qquad \lim_{h \to 0} \left(\frac{\sin h}{h}\cdot \frac{-\sin h}{\cos h+1}\right).$$

Finally, let $h \to 0$. We are allowed to use the fact that $\lim_{h\to 0}\frac{\sin h}{h}=1$. And it is clear that $\lim_{h\to 0}\frac{-\sin h}{\cos h +1}=0$. So our limit is $0$.

Comment: The idea used above is related to the process of "rationalizing the numerator," which you may have seen already, for example in finding the derivative of $\sqrt{x}$ at $x=3$ from the definition of the derivative.

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