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there is a question:

$$ \lim_{x->0}{\frac{x-arcsinx}{x^3(sinx+1)}} = \lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}} $$

[1]you know there have four place x. we know in limit calculation, if numerator x tend to 0, the denominator x must tend to 0 too, so can get the correct result.

but in the given question, there divided into two lim's multiplication:$$\lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}}$$

there I named them as lim_A· lim_B.

in my opinion, the lim_A and lim_B comes from one entirety, if lim_A's x tends to 0, the lim_B's x should tends to 0 too.

but however, in the question's solution there only let lim_A's x ->0 ignore the lim_B's x,

$$\lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}} = \lim_{x->0}{\frac{x-arcsinx}{x^3}}$$

so, why ? is there any theorem state this step? (I don't understand because I think the lim_A and lim_B comes from one entity, if one x->0, the others(lim_B's x) should x->0 too.)

244boy
  • 119

2 Answers2

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You may also do it as $$L=\lim_{x \to 0} \frac{x-\sin^{-1}x}{x^3}$$ by letting $x=\sin t$ as $$L=\lim_{t\to 0} \frac{\sin t- t}{\sin^3 t}= \lim_{t \to 0} \frac{1-t^3/3!+t^4/5!-...-t}{(t-t^3/3!+...)^3}=-\frac{1}{6}.$$

Dharmendra Singh
  • 987
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Note: your English or typing needs help! The word is opinion,not option.However there is no reason why both limits (not limitations) should $\to 0$.

For lim_B, get the first two terms in the power series for $arcsin(x)=x+\frac{x^3}{6}+...$ The answer is lim_B$=-\frac{1}{6}$

herb steinberg
  • 12,765