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Anamaria
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Este cumva 1 raspunsul?
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d
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Nu, este 1/2.
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Anamaria
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Pentru n suficient de mare,
Riguros, folosiți definiția cu epsilon a limitei
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[Citat]
Pentru n suficient de mare,
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Exact așa am făcut, de mi-a dat 1/2, dar nu mi-a plăcut cum am redactat.
Sper să fiu în stare să scriu o soluție rezonabila cu epsilon.
Mulțumesc mult!
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Anamaria
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Using the inequality
So
Using squeeze theorem, we have
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dtiwari
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Excellent idea.
ADMIN EDIT:
The above was corrected, although there is also a corrected form above.
Some notes: In such cases it may be simpler to write everything in a (one and only one) [eq uation] block.
By using $$ ... $$ in LaTeX, it is already passing to displaystyle.
\rightarrow can be simpler typed as \to
The latex compile error was in k^3^3, no double upper indices allowed, instead grouping an inner token as in {k^3}^3 works, but looks odd, (k^3)^3 is better. In our case, we need only k^3.
\frac{1}{2} also works as \frac 12 (or even \frac12). It only saves typing.
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dtiwari
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[Citat]
[Citat]
Using the inequality
So (eroare: eq.1/59172)$$\displaystyle \lim_{n\rightarrow\infty}\sigma^{n}_{k=1}d\frac{k}{n^2}<\tan \frac{k}{n^2}<\lim_{n\rightarrow \infty}\sigma^{n}_{k=1}\left(\dfrac{k}{n^2}+\frac{k^3^3}{3n^3}\right)$
Using squeeze theorem, we have (eroare: eq.2/59172)$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\tan\frac{k}{n^2} = \frac{1}{2}.$
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Raspunsul este 1.
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[Citat]
Raspunsul este 1. |
Este 1/2.
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Anamaria
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