Matematické Fórum

stuart clark · 25. 01. 2012 18:52

$\lim_{x\rightarrow 0}\frac{\arctan (x)-\arcsin (x)}{\sin^3 x}$

stuart clark · 25. 01. 2012 19:40

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Is there is any other method other then L, Hospital

If yes the how can i calculate it

Thanks

halogan · 25. 01. 2012 20:00

↑ stuart clark:

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stuart clark · 25. 01. 2012 20:12

Thanks ↑ halogan:

But It is given that without using Series expansion and L, Hospital Rule

stuart clark · 26. 01. 2012 09:43

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Pavel Brožek

↑ stuart clark:

Usually when you use algebraic limit theorem, you don't have to prove that the limits exist, because in every step you only separate one limit to more simpler limits. Then you show that the simpler limits exist and it is obvious that (by changing the direction of computation) you used the algebraic limit theorem properly. But here you separate one limit to two and later you put the two limits back together. I don't see why $\lim_{t\rightarrow 0}\left(\frac{t-\tan t}{t^3}\right)$ and $\lim_{t\rightarrow 0}\left(\frac{t-\sin t}{t^3}\right)$ should exist and be finite (of course it is obvious from the Taylor series of sin and tan, but I think that you wanted to avoid using Taylor series).

For example:

$0&=\lim_{x\to0}0=\lim_{x\to0}$\frac1{\sin^2x}-\frac1{\sin^2x}$=\lim_{x\to0}\frac1{\sin^2x}-\lim_{x\to0}\frac1{\sin^2x}=\\ &=\lim_{x\to0}\frac1{\sin^2x}-\lim_{x\to0}$\frac{\tan^2x}{\sin^2x}\cdot\frac1{\tan^2x}$=\\ &=\lim_{x\to0}\frac1{\sin^2x}-1\cdot\lim_{x\to0}\frac1{\tan^2x}=\\ &=\lim_{x\to0}$\frac1{\sin^2x}-\frac1{\tan^2x}$=\\ &=\lim_{x\to0}\frac{1-\cos^2x}{\sin^2x}=\\ &=\lim_{x\to0}1=1$

This "proof of 0=1" is not correct because of the indeterminate form $\infty-\infty$ .