$$\lim _{ { x }\to { 0 } }{ \frac { \sin x-\arctan x }{ {x }^{ 2 }\log(1+x) } }$$
this log is natural logarithm
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2If you know the power series for $\sin x$, $\arctan x$, and $\log (1+x)$, then you can apply those, otherwise, this is just a painful L'Hopital's rule problem. – Thomas Andrews Jul 08 '13 at 16:47
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A related problem. – Mhenni Benghorbal Jul 08 '13 at 19:54
1 Answers
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Using Series Expansion
$$\frac{\sin x-\arctan x}{x^2\log(1+x)}$$
$$=\frac{x-\frac {x^3}{3!}+O(x^5)-\{x-\frac {x^3}3++O(x^5)\}}{x^3}\cdot\frac1{\frac{\ln(1+x)}x}$$
$$=\frac{x^3(\frac13-\frac16)+O(x^5)}{x^3}\cdot\frac1{\frac{\ln(1+x)}x}$$
$$=\{\frac16+O(x^2)\}\cdot\frac1{\frac{\ln(1+x)}x}\text{ if }x\ne0$$
If $x\to0,x\ne0$
lab bhattacharjee
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i.e., 1/6. You could also find this with WolframAlpha, where it gives a step-by-step solution. – Benjamin Dickman Jul 08 '13 at 16:47
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