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Find $\displaystyle \lim_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}$

I tried using L'hopital's rule but it got very messy very fast


UPDATE- So reading about the Taylor series this is what I have so far

$$\lim_{x\to0}\frac{(x-x^3/3!+x^5/5!-\cdots)-(x+x^3/3+2x^5/5+17x^7/315+\cdots)}{(x+x^2/6+3x^5/40+\cdots)-(x-x^3/3+x^5/5-\cdots)}$$

But I'm still stuck

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marked as duplicate by Arnaud D., Carl Mummert, max_zorn, Delta-u, GNUSupporter 8964民主女神 地下教會 Oct 8 '18 at 10:12

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You can evaluate this limit using Taylor's expansions. I will write the expansions of the functions below. Note that you can ignore higher powers of x as it is tending to zero. (You will need to think of how many terms to ignore and how many you shouldn't). If you are still not able to evaluate, ask me for help.

\begin{align} \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} + \dotsb \\[4px] \tan x &= x + \frac{x^3}{3} + \dotsb \\[4px] \arcsin x &= x +\frac{1^2}{3!}x^3 + \frac{1^23^2}{5!}x^5 + \dotsb \\[4px] \arctan x &= x - \frac{x^3}{3} + \dotsb \end{align} You can search for more expansions and how to make these expansions in detail on the net or stack exchange itself.

here is the solution

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  • $\begingroup$ do we just arbitrarily say "oh let's go to the fifth term because it is accurate enough" and then calculate? $\endgroup$ – John Rawls Oct 3 '18 at 4:47
  • $\begingroup$ Evaluating a limit means to bring the given expression from an indeterminate form (here it is 0 divided by 0) to a determinate one. Your aim to evaluate this limit is bringing the expression out of the indeterminate form without changing its value. so when using expansions, you try to use minimum no. Of terms to bring the expression in a determinate one and ignore the rest if they are tending to zero.... $\endgroup$ – Math Maniac Oct 3 '18 at 4:52
  • $\begingroup$ okay for right now i bascially canceled out the x+x^3+x^5... and factored it out and got a fraction with a lot of messy numbers. what do i do in solving this? $\endgroup$ – John Rawls Oct 3 '18 at 4:54
  • $\begingroup$ Try to use only first two terms in expansion in this question... Wait a minute i will send the solution $\endgroup$ – Math Maniac Oct 3 '18 at 4:55
  • $\begingroup$ I edited my ans (posted a photo). Check it out $\endgroup$ – Math Maniac Oct 3 '18 at 4:58
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You were on the right track $$A=\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}=\frac{\left(x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)\right)-\left(x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right) \right)}{\left(x+\frac{x^3}{6}+\frac{3 x^5}{40}+O\left(x^7\right)\right)-\left(x-\frac{x^3}{3}+\frac{x^5}{5}+O\left(x^7\right) \right)}$$ Simplify to get $$A=\frac{-\frac{x^3}{2}-\frac{x^5}{8}+O\left(x^7\right)} {\frac{x^3}{2}-\frac{x^5}{8}+O\left(x^7\right) }=-1-\frac{x^2}{2}+O\left(x^4\right)$$ which shows the limit and how it is approached.

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This is a nice exercise which shows that a lot more can be achieved using standard limits than what most beginners would think.

We are going to use the following standard limits $$\lim_{x\to 0} \frac{\sin x} {x} =1,\lim_{x\to 0} \frac{1-\cos x} {x^2}=\frac{1}{2},\lim_{x\to 0}\frac{\arctan x} {x} =1$$ All of these are immediate consequences of the first limit.

The numerator of the given expression can be written as $$\frac{\sin x} {x} \cdot x^3\cdot\frac{\cos x-1}{x^2}\cdot\frac{1}{\cos x} $$ and using the standard limit we can replace the above with $-x^3/2$.

The denominator needs a little more effort. Using the identities $$\arcsin x=\arctan\frac{x} {\sqrt{1-x^2}},\arctan x-\arctan y=\arctan\frac{x-y} {1+xy}$$ we can write the denominator as $$\arctan\dfrac{\dfrac{x} {\sqrt{1-x^2}}-x } {1+\dfrac{x^2}{\sqrt{1-x^2}}} $$ and using the limit $\lim_{x\to 0}(\arctan x) /x=1$ the above expression can be replaced by $$\dfrac{\dfrac{x} {\sqrt{1-x^2}}-x } {1+\dfrac{x^2}{\sqrt{1-x^2}}}=\frac{x(1-\sqrt{1-x^2})}{x^2+\sqrt{1-x^2}}$$ Thus the desired limit is equal to the limit of $$\dfrac{-x^3/2}{\dfrac{x(1-\sqrt{1-x^2})}{x^2+\sqrt{1-x^2}}}=-\frac{x^2(x^2+\sqrt{1-x^2})}{2(1-\sqrt{1-x^2})}$$ Multiplying numerator and denominator by $1+\sqrt{1-x^2}$ we can simply the expression above as $$-\frac{(x^2+\sqrt{1-x^2})(1+\sqrt {1-x^2})} {2} $$ and this clearly has limit $-1$ as $x\to 0$.

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