| cos√2(x+h)−cos√2x | ||
limh→0 | = | |
| h |
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limh→0 | = | ||||||||||||||||
| h |
| √2x+√2x |
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−2sin | *limh→0 | = | |||||||||||
| 2 | h |
| √2(x+h)−√2x |
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−2sin√2x*limh→0 | * | = | |||||||||||
| 2h |
|
| (√2x+h−√2x)(√2x+h+√2x) | ||
−2sin√2x*1*limh→0 | ||
| h*(√2x+h+√2x) |
| 2x+h−2x | ||
−2sin√2x*limh→0 | = | |
| h*(√2x+h+√2x) |
| h | ||
−2sin√2x*limh→0 | = | |
| h*(√2x+h+√2x) |
| 1 | ||
−2sin√2x*limh→0 | = | |
| √2x+h+√2x |
| 1 | −2sin√2x | sin√2x | ||||
−2sin√2x* | = | = − | ||||
| √2x+√2x | 2√2x | √2x |
| sinx | ||
zakładam, że już udowodniliście, że limx→0 | =1 | |
| x |