Evaluar el límite $\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$

Question

Necesito evaluar el límite sin usar regla de l'Hopital.

$$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$

Answer 1

$\frac{\sqrt{1+x\sin(x)}-\sqrt{\cos(2x)}}{\tan^{2}(\frac{x}{2})}=\bigg(\frac{1-\cos(2x)}{\tan^{2}(\frac{x}{2})}+\frac{x\sin(x)}{\tan^{2}(\frac{x}{2})}\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}=\bigg(\frac{1-\cos(2x)}{(2x)^{2}}\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}(16\cos^{2}(\frac{x}{2}))+\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}\frac{\sin(x)}{x}(4\cos^{2}(\frac{x}{2}))\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}$

Dejar $x$ vaya a $0$ en la última formulación del límite da:

$(\frac{1}{2}\cdot1\cdot16+1\cdot1\cdot4)\frac{1}{2}=6$.

Answer 2

Si se nos permite usar serie de Taylor:

Desde $\sin x = x + O(x^3)$, tenemos $1+x\sin x = 1 + x^2 + O(x^4)$.

Así, $\sqrt{1+x\sin x} = 1+\dfrac{1}{2}x^2 + O(x^4)$.

Desde $\cos x = 1 - \dfrac{1}{2}x^2 + O(x^4)$, tenemos $\cos 2x = 1 - 2x^2 + O(x^4)$.

Así, $\sqrt{\cos 2x} = 1-x^2 + O(x^4)$.

Desde $\tan x = x + O(x^3)$, tenemos $\tan \dfrac{x}{2} = \dfrac{1}{2}x + O(x^3)$.

Así, $\tan^2 \dfrac{x}{2} = \dfrac{1}{4}x^2 + O(x^4)$.

Por lo tanto, $\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\tan^2\tfrac{x}{2}} = \dfrac{\left(1+\tfrac{1}{2}x^2 + O(x^4)\right) - \left(1-x^2+O(x^4)\right)}{\tfrac{1}{4}x^2 + O(x^4)} = \dfrac{\tfrac{3}{2}x^2+O(x^4)}{\tfrac{1}{4}x^2+O(x^4)} = \dfrac{\tfrac{3}{2}+O(x^2)}{\tfrac{1}{4}+O(x^2)}$.

Por lo tanto, como $x \to 0$, $\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\tan^2\tfrac{x}{2}} \to \dfrac{\tfrac{3}{2}}{\tfrac{1}{4}} = 6$ %.