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Hernaner28
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Hi. I just want to ask: how can I realize that I need to do the 4th order taylor's expansion for solving a precise limit? e.g.
[tex]\mathop {\lim }\limits_{x\to 0} \frac{{{e^x}-1-\frac{{{x^2}}}{2}+\sin x-2x}}{{1-\cos x-\frac{{{x^2}}}{2}}}[/tex]
We need the 4th order of the expansion but how can I realize just seeing it?another question. Why is this true:
[tex]\displaystyle\mathop {\lim }\limits_{x \to 0} \frac{{o(x)}}{x} = 0[/tex]?
o(x) is the remainder (Landau notation)
Thanks!
[tex]\mathop {\lim }\limits_{x\to 0} \frac{{{e^x}-1-\frac{{{x^2}}}{2}+\sin x-2x}}{{1-\cos x-\frac{{{x^2}}}{2}}}[/tex]
We need the 4th order of the expansion but how can I realize just seeing it?another question. Why is this true:
[tex]\displaystyle\mathop {\lim }\limits_{x \to 0} \frac{{o(x)}}{x} = 0[/tex]?
o(x) is the remainder (Landau notation)
Thanks!
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