h→0limhsin(h−hcos(h)) So, we know that
h→0limhsin(h)=1 Let's multiply the numerator and denominator by (h-hcos(h)):
h→0limh∗(h−hcos(h))sin(h−hcos(h))∗(h−hcos(h))=(1)h→0limh−hcos(h)sin(h−hcos(h))=1
(1)=h→0limhh−hcos(h)=h→0lim1−cos(h)Use the Maclaurin series for cosine:
cos(h)=1−2!h2+4!h4−6!h6+...
1−cos(h)=1−(1−2!h2+...)=2h2+...=O(h2) Answer: the function has quadratic convergence.
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