Calcular $\lim_{\theta\rightarrow 0}\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}$

Question

Reescribí por escrito el bronceado como sin/cos y la cruz de la multiplicación:

$$\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}= \frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\frac{\sin{(\tan{\theta})}\cos{(\sin{\theta})-\cos{(\tan{\theta})\sin{(\sin{\theta})}}}}{\cos{(\tan{\theta})\cos{(\sin{\theta})}}}}.$$

Mediante la suma de la fórmula para el seno de recibir: $$\sin{(\tan{\theta})}\cos{(\sin{\theta})-\cos{(\tan{\theta})\sin{(\sin{\theta})}}}=\sin{(\tan{\theta}}-\sin{\theta}), $$

Desde $\cos{(\tan{\theta})}\cos{(\sin{\theta})}\rightarrow1$ $\theta\rightarrow 0,$ el problema se reduce a encontrar el límite de $$\lim_{\theta\rightarrow0}\frac{\sin{(\tan{\theta})-\sin{(\sin{\theta})}}}{\sin{(\tan{\theta}-\sin{\theta})}}.$$

Answer 1

$$\begin{align}&\lim_{\theta\rightarrow0}\frac{\sin{(\tan{\theta})-\sin{(\sin{\theta})}}}{\sin{(\tan{\theta}-\sin{\theta})}} &\\=& \lim_{\theta\rightarrow0}\frac{2 \sin\left({\dfrac{\tan{\theta} -\sin{\theta}}2}\right)\cos\left({\dfrac{\tan{\theta} +\sin{\theta}}2 } \right)}{2\sin{\left(\dfrac{\tan{\theta}-\sin{\theta}}{2}\right)}\cos{\left(\dfrac{\tan{\theta}-\sin{\theta}}{2}\right)}}&\\ =&\lim_{\theta\rightarrow0}\frac{\cos\left({\dfrac{\tan{\theta} +\sin{\theta}}2 } \right)}{\cos{\left(\dfrac{\tan{\theta}-\sin{\theta}}{2}\right)}} = 1\end{align}$$

Answer 2

Utilizando el Valor medio Teorema, $$ \begin{align} \lim_{\theta\to0}\frac{\sin(\tan(\theta))-\sin(\sin(\theta))}{\tan(\tan(\theta))-\tan(\sin(\theta))} &=\lim_{\theta\to0}\frac{\frac{\sin(\tan(\theta))-\sin(\sin(\theta))}{\tan(\theta)-\sin(\theta)}}{\frac{\tan(\tan(\theta))-\tan(\sin(\theta))}{\tan(\theta)-\sin(\theta)}}\\ &=\lim_{\theta\to0}\frac{\cos(\xi_1(\theta))}{\sec^2(\xi_2(\theta))}\\[6pt] &=\frac{\cos(0)}{\sec^2(0)}\\[12pt] &=1 \end{align} $$ donde $\xi_1(\theta)$ $\xi_2(\theta)$ entre $\sin(\theta)$$\tan(\theta)$.

Answer 3

Deje $x = \sin \theta$. Entonces

$$ \lim_{\theta\rightarrow 0}\frac{\sin{(\bronceado{\theta})}-\sin{(\sin{\theta})}}{\bronceado{(\bronceado{\theta})}-\bronceado{(\sin{\theta})}}\\ = \lim_{x\rightarrow 0}\frac{\sin{(\frac{x}{\sqrt{1 - x^2}})}-\sin{(x)}}{\bronceado{(\frac{x}{\sqrt{1 - x^2}})}-\bronceado{(x)}}\\ $$

Además, por trig teoremas (por ejemplo, en la wikipedia) ,

$\sin a - \sin b = 2 \cos ((a+b)/2) \sin ((a-b)/2)$

y

$\tan a - \tan b = \frac{\sin (a-b) }{\cos a \cos b} = \frac{2 \sin ((a-b)/2)\cos ((a-b)/2) }{\cos a \cos b} = $

Así, obtenemos

$\frac{\pecado - \pecado b}{\bronceado a - \bronceado b} = \frac{ \cos ((a+b)/2) \cos \cos b}{\cos ((a-b)/2) } $

Así tenemos $$ \lim_{x\rightarrow 0}\frac{\cos{((\frac{x}{\sqrt{1 - x^2}} + x)/2)}\cos{(x)}\cos{(\frac{x}{\sqrt{1 - x^2}})}}{\cos{((\frac{x}{\sqrt{1 - x^2}} - x)/2)}} = 1 $$

puesto que todas las funciones angulares dar $\cos 0 = 1$