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Looked around a bit and all I see are proofs using the limit definition of a derivative. This is not for an assignment, I could just use the limit definition if I wanted to, but I was wondering how you could go about proving this using the epsilon-delta definition of a derivative ($\forall \epsilon >0$, $\exists \delta >0$ such that if $0< |x-c| <\delta$, then $\left | \frac{f(x) - f(c)}{x-c} - f'(c) \right | < \epsilon$).

Edit: Just to be clear I am explicitly looking for an epsilon-delta formulation of the proof. Was just trying to prove this without throwing a bunch of trig identities and limit theorems at the problem.

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    $\begingroup$ It's useful to write $\sin(x+h) = \sin x \cos h + \sin h \cos x$. $\endgroup$ – Elchanan Solomon Feb 18 '13 at 3:22
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    $\begingroup$ What do you believe the definitions of $\sin x$ and $\cos x$ are? $\endgroup$ – Micah Feb 18 '13 at 4:08
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I know this post is somewhat old, but I'll post a thorough answer anyway so it may benefit future readers.

To prove this, we will want to use the limits $\frac {\sin x}x \to 1$ and $\frac {1 - \cos x}x \to 0$ when $x \to 0$. By the $\epsilon-\delta$ definition, we can write (quantifiers left out, assuming reader is familiar with the definition):

$$ (1): 0 < |x - 0| < \delta_1 \implies \left|\frac {\sin x}x - 1\right| < \frac {\epsilon}2 \\ (2): 0 < |x - 0| < \delta_2 \implies \left|\frac {1 - \cos x}x - 0\right| < \frac {\epsilon}2 \\$$

Now let $\delta = \min \{ \delta_1, \delta_2 \}$. What we then want to show is that

$$ 0 < |x - 0| < \delta \implies \left|\frac {\sin(x+h) - \sin x}x - \cos x\right| <\epsilon $$

Using $\sin (x +h) = \sin x \cos h + \sin h \cos x$ and some algebra inside the absolute value sign, we can write

$$ \left|\frac {\sin(x+h) - \sin x}x - \cos x\right| = \left| \frac { \sin x (\cos h - 1) }h + \frac { \cos x (\sin h - h) }h \right| $$

By the triangle inequality, we have

$$\left| \frac { \sin x (\cos h - 1) }h + \frac { \cos x (\sin h - h) }h \right| \leq | \sin x | \left| \frac {\cos h - 1}h \right| + | \cos x | \left| \frac {\sin h - h}h \right|$$

Trivially, $|\sin x | \leq 1$ and $| \cos x | \leq 1$. Therefore

$$ | \sin x | \left| \frac {\cos h - 1}h \right| + | \cos x | \left| \frac {\sin h - h}h \right| \leq \left| \frac {\cos h - 1}h \right| + \left| \frac {\sin h - h}h \right| $$

But by $(1)$ and $(2)$ we finally have

$$ \left| \frac {\cos h - 1}h \right| + \left| \frac {\sin h - h}h \right| < \frac {\epsilon}2 + \frac {\epsilon}2 = \epsilon $$

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It may depend on what limits you know, but

$$\frac{\sin(x+h)-\sin x}{h}=\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}=$$

$$=\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\xrightarrow[h\to 0]{}\sin x\cdot 0+\cos x\cdot 1 =\cos x$$

You can also try other trigonometric identities with an equivalent definition fro the derivative:

$$\frac{\sin x-\sin x_0}{x-x_0}=2\frac{\sin\frac{x-x_0}{2}\cos\frac{x+x_0}{2}}{x-x_0}=$$

$$=\frac{\sin\frac{x-x_0}{2}}{\frac{x-x_0}{2}}\cos\frac{x+x_0}{2}\xrightarrow[x\to x_0]{}1\cdot \cos\frac{2x_0}{2}=\cos x_0$$

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    $\begingroup$ OP was explicitly asking for an $\epsilon$-$\delta$-formulation. $\endgroup$ – Martin Feb 18 '13 at 3:45
  • $\begingroup$ So he was...I missed that. Oh, well: there still is some stuff for others to see. $\endgroup$ – DonAntonio Feb 18 '13 at 3:55
  • $\begingroup$ All the usual proofs seem to depend on $sin'(0) = 1$; i.e., $\lim_{h \to 0} sin(h)/h = 1$. $\endgroup$ – marty cohen Feb 18 '13 at 4:43
  • $\begingroup$ The first one also depends on $\,\frac{\cos h-1}{h}\to 0=\cos'(0)\,$. This is why I remarked this depends on what limits the OP knows... $\endgroup$ – DonAntonio Feb 18 '13 at 4:45

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