MHB Evaluate the limit of a trigonometric expression

[lfdahl]

Evaluate:

\[\lim_{x\rightarrow 0}\frac{\sin (\arctan x)-\tan (\arcsin x)}{\arcsin (\tan x)-\arctan (\sin x)}\]

 

[Theia]

Spoiler

Something like this(?):

Assume $$|x| \approx 0$$. Then

$$\begin{align*} \sin (\arctan x)-\tan (\arcsin x) &= \frac{x}{\sqrt{1 + x^2}} - \frac{x}{\sqrt{1-x^2}} \\
&= x\left( 1 - \frac{x^2}{2} + \cdots \right) - x\left( 1 + \frac{x^2}{2} \right) \\
&= -x^3 + \cdots\end{align*}$$

and

$$\begin{align*} \arcsin (\tan x) - \arctan (\sin x) &= \tan x + \frac{\tan^3 x}{6} + \cdots - \sin x + \frac{\sin^3 x}{3} + \cdots \\
&= x + \frac{x^3}{3} + \cdots +\frac{x^3}{6} + \cdots - \left( x - \frac{x^3}{6} + \cdots \right) + \frac{x^3}{3} + \cdots \\
&= x^3 + \cdots\end{align*}$$.

Hence

\[\lim_{x\rightarrow 0}\frac{\sin (\arctan x)-\tan (\arcsin x)}{\arcsin (\tan x)-\arctan (\sin x)} = -1.\]

 

[lfdahl]

Spoiler

Something like this(?):

Assume $$|x| \approx 0$$. Then

$$\begin{align*} \sin (\arctan x)-\tan (\arcsin x) &= \frac{x}{\sqrt{1 + x^2}} - \frac{x}{\sqrt{1-x^2}} \\
&= x\left( 1 - \frac{x^2}{2} + \cdots \right) - x\left( 1 + \frac{x^2}{2} \right) \\
&= -x^3 + \cdots\end{align*}$$

and

$$\begin{align*} \arcsin (\tan x) - \arctan (\sin x) &= \tan x + \frac{\tan^3 x}{6} + \cdots - \sin x + \frac{\sin^3 x}{3} + \cdots \\
&= x + \frac{x^3}{3} + \cdots +\frac{x^3}{6} + \cdots - \left( x - \frac{x^3}{6} + \cdots \right) + \frac{x^3}{3} + \cdots \\
&= x^3 + \cdots\end{align*}$$.

Hence

\[\lim_{x\rightarrow 0}\frac{\sin (\arctan x)-\tan (\arcsin x)}{\arcsin (\tan x)-\arctan (\sin x)} = -1.\]

Hi, Theia! Thankyou for a nice solution and your participation!(Yes)