# 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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It should say $f(0 + \Delta x) - f(0)$ in the first line. You want to show that the difference in function values over a small interval $\Delta x$, divided by the width of this interval (i.e. the "average" increase in $f$ on this interval) converges to some value $L = f'(0)$ as $\Delta x \to 0$. Also, to get something useful, you should first write out the formula with $\Delta x$, and only when you get something clean enough, let $\Delta x \to 0$. You lost the limit too soon and did not simplify anything. – TMM Sep 11 '11 at 19:08
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

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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

## 2 Answers

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

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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
– Srivatsan Sep 11 '11 at 19:42
@Sri: Ah, you're right. I wasn't sure how you'd go from e.g. $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$ and others to $\cos(2x) = 1 - 2\sin^2(x)$, but it's quite easy after all ;) – TMM Sep 11 '11 at 19:55
@BKaylor: One limit is enough, no need for two! So you can use $\cos'(0) = \lim_{\Delta x \to 0} \frac{\cos(0+\Delta x) - \cos(0)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\cos(\Delta x) - 1}{\Delta x} = 0$ and you are done. – TMM Sep 11 '11 at 19:57

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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