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The question is how to show the derivative of $\cos x$ is $-\sin x$ using the definition of the derivative.

I do this proof in the normal way by using the sum of $\cos (x+h)$ using the trig identity, and then factoring out the $\cos x$ and using two special squeeze theorems. I get the correct answer of $-\sin x$.

Here is how he does it and it seems illegal:

$$\lim_{h \mapsto 0 } \frac{(\cos x)(\cos h)-(\sin x)(\sin h) -\cos x}h $$ then he says in the NUMERATOR, letting $ h=0$ makes the numerator become:

$\cos x (1)-\sin x\sin h-\cos x$ Notice that he just selectively let the $\cos h$ term have $h$ go to zero but NOT the $\sin h$ term!!!

The two $\cos x $ terms cancel out leaving $\lim_{h \mapsto 0 }-\sin x\sin h/h$ and he uses the squeeze theorm to get $-\sin x (1)$

But I don't see how it is legal to let $h=0$ for some terms in the numerator without considering the denominator. He DOES get the right answer, but it seems illegal to me. Thoughts?

Faraad Armwood
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user163862
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3 Answers3

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\begin{align} & \lim_{h\to0}\frac{(\cos x)(\cos h)-(\sin x)(\sin h) -\cos x}h \\[12pt] = {} & (\cos x)\,\, \underbrace{\lim_{h\to0} \left( \frac{(\cos h) -1} h \right)}_\text{first limit} - (\sin x)\,\, \underbrace{\lim_{h\to0} \left( \frac{\sin h} h \right)}_\text{second limit} \end{align} The first limit is $0$ and the second is $1$.

That's the way to do this. He shouldn't be letting $h=0$ in the numerator while leaving $h$ in the denominator. The fact that the first limit is $0$ cannot be shown merely by showing that the limit of the numerator is $0.$ If that were valid, the the same argument argument would show that the second limit is $0,$ and that is wrong.

Michael Hardy
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  • How do you calculate the first? – Nosrati Aug 26 '17 at 19:51
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    @MyGlasses: Multiply and divide by $\cos(h)+1$. – Ted Shifrin Aug 26 '17 at 19:52
  • @MyGlasses It is a standard limit, but here's a derivation: $$\lim_{x\to0} \frac{\cos(x)-1}{x} = \lim_{x\to0} \frac{\cos^2(x)-1}{x(\cos(x)+1)} = \lim_{x\to0} \frac{\sin^2(x)}{x(\cos(x)+1)} \ = \lim_{x\to0} \frac{\sin(x)}{x} \cdot \lim_{x\to0} \frac{1}{\cos(x)+1} \cdot \lim_{x\to0} \sin(x) = 1 \cdot \frac12 \cdot 0$$ – wythagoras Aug 26 '17 at 19:52
  • @MyGlasses : I'll be back to address that later; right now I must attend to other things. I think I can write a far simpler argument that that of "wpythagoras". – Michael Hardy Aug 26 '17 at 19:53
  • @wythagoras: There's an unfortunate and confusing line break there. At least put a \cdot at the end of the line? Better, move that term down to the second line. – Ted Shifrin Aug 26 '17 at 19:53
  • Following on @MichaelHardy's "far simpler" suggestion, try the double angle formula for $\cos(2\theta)-1$. I'm not sure I agree with "far simpler," but it's an alternative derivation. – Ted Shifrin Aug 26 '17 at 19:54
  • For the first limit, you have $\cfrac {-2\sin^2 \frac h2}h$ – Mark Bennet Aug 26 '17 at 19:54
  • and $\cos^2x-1=-\sin^2x$ – Nosrati Aug 26 '17 at 19:55
  • @MyGlasses : Picture the unit circle $a^2+b^2=1$ being about a mile in radius and you're standing on the $a$-axis next to the point $(a,b) = (1,0),$ and $h$ is about $10$ feet. Then $\cos h$ is the length of the horizontal line from the $b$-axis to a point about $10$ feet above your feet. Where you're standing the circle looks nearly like a vertical wall. So $1-\cos h$ is the tiny distance from where the wall would have been at $10$ feet above the ground if it were perfectly vertical, and where it actually is at that point. That should show you why $1-\cos h$ is next to$,\ldots\qquad$ – Michael Hardy Aug 26 '17 at 20:56
  • $\ldots,$nothing by comparison to $h$ when $h$ is small. $\qquad$ – Michael Hardy Aug 26 '17 at 20:56
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Maybe your teacher isn't phrasing it carefully or well enough. What we do is write the limit as

$$ \lim_{h \to 0} \frac{(\cos x)(\cos h) - (\sin x)(\sin h) - \cos x}{h} = \lim_{h \to 0} \left[ \cos x \left( \frac{\cos h - 1}{h} \right) - \sin x \left( \frac{\sin h}{h} \right) \right] $$

and we use our limit rules:

$$ \lim_{h \to 0} \left[ f(h) - g(h) \right] = \left[ \lim_{h \to 0} f(h) \right] - \left[ \lim_{h \to 0} g(h) \right] $$

provided the limits exist.

Thus

$$ \hspace{-2cm} \lim_{h \to 0} \left[ \cos x \left( \frac{\cos h - 1}{h} \right) - \sin x \left( \frac{\sin h}{h} \right) \right] \\ \qquad= \left[ \lim_{h \to 0} \cos x \left( \frac{\cos h - 1}{h} \right) \right] - \left[ \lim_{h \to 0} \sin x \left( \frac{\sin h}{h} \right) \right] \\ \qquad = \left[ \cos x \lim_{h \to 0} \left( \frac{\cos h - 1}{h} \right) \right]-\left[ \sin x \lim_{h \to 0} \left( \frac{\sin h}{h} \right) \right] $$

And now we use the limits

$$ \lim_{h \to 0} \left( \frac{\cos h - 1}{h} \right) = 0 \text{ and } \lim_{h \to 0} \left( \frac{\sin h}{h} \right) = 1 $$

to get

$$ \lim_{h \to 0} \left[ \cos x \left( \frac{\cos h - 1}{h} \right) - \sin x \left( \frac{\sin h}{h} \right) \right] = - \sin x. $$

Trevor Gunn
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I respect to other solutions but I prefer this $$\lim_{h\to0}\dfrac{\cos(x+h)-\cos(x)}{h}=\lim_{h\to0}\dfrac{-2\sin(x+\frac{h}{2})\sin(\frac{h}{2})}{h}=-\sin(x)$$

Nosrati
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