Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
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$\begingroup$ Sorry for mistake, I have edited a problem but thanks for replies. $\endgroup$ – user300045 Jun 6 '15 at 23:01
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1$\begingroup$ So you posted your question incorrectly, and after three persons wrote a reply, you decided to edit your question. That is not respectful behaviour. $\endgroup$ – M. Wind Jun 6 '15 at 23:16
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1$\begingroup$ Maybe op only just logged back on? People are pretty quick to respond with answers here! $\endgroup$ – danimal Jun 6 '15 at 23:35
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Let $y=\arctan x$. This uses two limits: $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$.
$$\lim_{x\to 0}{\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arctan(x)}\right)}$$
$$=\lim_{x\to 0}\frac{1-\cos x}{\sin x\arctan x}=\lim_{x\to 0}\left(\frac{1-\cos x}{x^2}\cdot \frac{x}{\sin x}\cdot \frac{x}{\arctan x}\right)$$
$$=\frac{1}{2}\cdot \lim_{y\to 0}\frac{\tan y}{y}=\frac{1}{2}\cdot \lim_{y\to 0}\frac{\sin y}{y}\cdot \lim_{y\to 0}\frac{1}{\cos y}=\frac{1}{2}$$
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Note $$\lim_{x\to0}\frac{\sin(x)}{x}=1, \lim_{x\to0}\frac{\arctan(x)}{x}=1,\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12, $$ and hence \begin{eqnarray} &&\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arctan(x)}\right)\\ &=&\lim\limits_{x\to\ 0}\frac{1-\cos x}{\sin(x)\arctan(x)}\\ &=&\lim\limits_{x\to\ 0}\frac{\frac{1-\cos x}{x^2}}{\frac{\sin(x)}{x}\frac{\arctan(x)}{x}}\\ &=&\frac12. \end{eqnarray}
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The series expansions are
$\sin x = x - x^3/6...,\,\,\tan x = x + x^3/3...,\,\, \arctan x=x - x^3/3...$
Equate numerators
$$\frac{(x+x^3/3)-(x-x^3/6)}{(x - x^3/6)(x + x^3/3)(x^3/3)}$$
$$\frac{x^3/2}{(x - x^3/6)(x + x^3/3)(x^3/3)}$$
We only care about the $x^3$ in the numerator
$$\frac{x^3/2}{x^3}$$
$$1/2$$
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Notice that the limit is
$$ L=\lim_{x \to 0}\frac{1}{\sin x} \left(\frac{1}{\arctan x} - \frac{\cos x}{\arcsin x} \right) $$ from the series expansion for small values of $x$ one has $\cos x =1- \frac{x^2}{2} + O(x^4)$ and $(\mbox{arc})\sin x = x + O(x^3) = \arctan x$ hence
$$L = \lim_{x \to 0}\frac{1}{(x + O(x^3))} \left(\frac{1}{x + O(x^3) } - \frac{1-x^2/2 + O(x^4)}{x + O(x^3)} \right)=\lim_{x \to 0} \frac{1}{x}\left(\frac{x}{2} + O(x^3) \right) = \frac{1}{2}$$
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With equivalence, it is faster (as often): $$1-\cos x\sim_0\frac{x^2}2,\enspace \sin x\sim_0 x,\enspace \arctan x\sim_0 x,\enspace\text{whence}\enspace\frac{1-\cos x}{\sin x\arctan x}\sim_0 \frac{x^2}{2x^2}=\frac12.$$
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Here is a solution to the EDITED problem asked by OP. I will make use of some standard limits only, without l'Hospital rule nor Taylor series. First, some simple transformations are required: \begin{eqnarray*} \frac{1}{\sin x\arctan x}-\frac{1}{\tan x\arcsin x} &=&\frac{1}{\sin x\arctan x}-\frac{\cos x}{\sin x\arcsin x} \\ &=&\frac{1}{\sin x}\left( \frac{1}{\arctan x}-\frac{\cos x}{\arcsin x}% \right) \\ &=&\frac{1}{\sin x}\left( \frac{\arcsin x-\cos x\arctan x}{\arctan x\arcsin x% }\right) . \end{eqnarray*} Now, we re-write this expression using the expressions involved in standard limits as follows: \begin{eqnarray*} &=&\frac{1}{\sin x}\left( \frac{\arcsin x-x+x-\arctan x+\arctan x-\cos x\arctan x}{\arctan x\arcsin x}\right) \\ &=&\frac{1}{\sin x}\left( \frac{\left( \arcsin x-x\right) +\left( x-\arctan x\right) +\arctan x\left( 1-\cos x\right) }{\arctan x\arcsin x}\right) \\ &=&\frac{x}{\sin x}\frac{x}{\arctan x}\frac{x}{\arcsin x}\left( \frac{\left( \arcsin x-x\right) +\left( x-\arctan x\right) +\arctan x\left( 1-\cos x\right) }{x^{3}}\right) \\ &=&\frac{x}{\sin x}\frac{x}{\arctan x}\frac{x}{\arcsin x}\left( \frac{% \arcsin x-x}{x^{3}}+\frac{x-\arctan x}{x^{3}}+\frac{\arctan x}{x}\left( \frac{1-\cos x}{x^{2}}\right) \right) \end{eqnarray*} Standard limits used are \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1,\ \ \ \ \lim_{x\rightarrow 0}% \frac{\arcsin x}{x}=1,\ \ \ \ \lim_{x\rightarrow 0}\frac{\arctan x}{x}=1 \\ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2},\ \ \ \ \lim_{x\rightarrow 0}\frac{\arcsin x-x}{x^{3}}=\frac{1}{6},\ \ \ \ \ \lim_{x\rightarrow 0}\frac{x-\arctan x}{x^{3}}=\frac{1}{3} \end{eqnarray*} Therefore, \begin{equation*} \lim_{x\rightarrow 0}\left( \frac{1}{\sin x\arctan x}-\frac{1}{\tan x\arcsin x}\right) =1\cdot 1\cdot 1\left( \frac{1}{6}+\frac{1}{3}+1\cdot \left( \frac{% 1}{2}\right) \right) =1. \end{equation*}
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$\begingroup$ +1. I like this, although I think it's reasonable to point out that your limits, in conjunction, are essentially equivalent to positing the first two non-zero terms of the Taylor series for $\sin$ and $\tan$. $\endgroup$ – Brian Tung Jun 8 '15 at 23:49
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The problem as corrected by the OP is now more symmetric. We can consider cases characterized as
$$ L = \lim_{x \to 0} \frac{1}{f(x)g^{-1}(x)}-\frac{1}{f^{-1}(x)g(x)} $$
where smooth $f(x) = x + \varepsilon_f x^m + o(x^{m+1}), g(x) = x + \varepsilon_g x^n + o(x^{n+1})$ with integers $m, n > 1$. Then $f^{-1}(x) = x - \varepsilon_f x^m + o(x^{m+1}), g^{-1}(x) = x - \varepsilon_g x^n + o(x^{n+1})$, and
$$ \begin{align} L & = \lim_{x \to 0} \frac{1}{(x+\varepsilon_fx^m)(x-\varepsilon_gx^n)} - \frac{1}{(x-\varepsilon_fx^m)(x+\varepsilon_gx^n)} \\ & = \lim_{x \to 0} \frac{2\varepsilon_gx^{n+1}-2\varepsilon_fx^{m+1}} {x^4+o(x^6)} \\ & = \lim_{x \to 0} 2\varepsilon_gx^{n-3}-2\varepsilon_fx^{m-3} \end{align} $$
In this case, we have $f(x) = \sin x, g(x) = \tan x$, so $\varepsilon_f = -1/6, \varepsilon_g = 1/3, m = n = 3$, and then
$$ L = \lim_{x \to 0} 2 \left(\frac{1}{3}\right) - 2 \left(-\frac{1}{6}\right) = 1 $$
