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Exercise 4

Avg difficulty 2.5

It can be shown that lim⁡h→0sin⁡(h)h=1\lim\limits_{h\to0}\frac{\fe{\sin}{h}}{h}=1 and lim⁡h→0cos⁡(h)−1h=0\lim\limits_{h\to0}\frac{\fe{\cos}{h}-1}{h}=0. Use these limits to help you to establish the first derivative formula for sin⁡(x)\fe{\sin}{x}.

Simonds, Calculus Lab Manual, §3.5 Supplement, Exercise 4

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Adapted from Calculus Lab Manual © 2012–2016 Portland Community College. Licensed under CC BY-SA 4.0. Source: Calculus Lab Manual. Imported at revision 3060f8e.