$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$

Question

Cálculo: $$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$ $

No sé cómo utilizar la regla de L'Hôpital.

He intentado hacer $\tan x =\frac{\sin x}{\cos x}$ el término ${\sqrt{\tan x}}$.

Answer 1

Primero puede quitar algunos factores $$ \lim{x \to 0 ^ +} \frac {\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}} {x ^ \sqrt{x}}\ 3 = \lim{x \to 0 ^ +} \frac {\tan x \cdot \sqrt{\tan x}} {x \sqrt{x}}\lim{x \to 0 ^ +} \frac {1-\cos x\sqrt {\cos x}} {x ^ 2} \ = \ lim {x \to 0 ^ +} \frac {1-\cos x\sqrt {\cos x}} {x ^ 2}. $$ y luego multiplicar por la conjugada $$=\lim{x \to 0^+}\frac{1-\cos^3 x}{x^2(1+\cos x\sqrt{\cos x})},$ $ evaluar el factor finito en el denominador $$=\frac12\lim{x \to 0^+}\frac{1-\cos x(1-\sin^2 x)}{x^2},$$ uso trigonométricas identitites $$=\frac12\lim_{x \to 0^+}\frac{2\sin^2\frac x2+\cos x\sin^2 x}{x^2},$ $ y concluir $$=\left(\frac12\right)^2+\frac12.$ $

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Hemos utilizado

$$\frac{\tan x}x=\frac{\sin x}x\frac1{\cos x}\to 1,$$ $$\frac{\sin ax}x=a\frac{\sin ax}{ax}\to a.$$

Answer 2

$$ \begin{align} \lim_{x\to0^{+}}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}} &=\lim_{x\to0^{+}}\frac{\tan^{3/2}x-\sin^{3/2} x}{x^3 \sqrt{x}}\\ &=\lim_{x\to0^{+}}\frac{\sin^{3/2}x}{x^{3/2}}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\left(\frac{\sin x}{x}\right)^{3/2}\cdot \lim_{n\to\infty}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\left(1+\frac12x^2+\frac1{24}x^4+\cdots\right)^{3/2}-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\left(1+\frac34x^2+\cdots\right)-1}{x^2}\\ &=\frac34 \end {align} $$

Answer 3

$$\frac{\tan x\sqrt{\tan x}-\sin x\sqrt{\sin x}}{x^3\sqrt x}=\left(\frac{\sin x} x\right)^{3/2}\cdot\frac{\frac1{\cos^{3/2}x}-1}{x^2}=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^{3/2}x}{x^2\cos^{3/2}x}=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^2x+\cos^{3/2}x(\cos^{1/2}x-1)}{x^2\cos^{3/2}x}=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2+\frac{\cos^{1/2}x-1}{x^2}\right]=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2+\frac{\cos x-1}{x^2(\cos^{1/2}x+1)}\right]=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2-\frac2{\cos^{1/2}x+1}\frac{\sin^2\frac x2}{x^2}\right]=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2-\frac2{\cos^{1/2}x+1}\cdot\frac14\cdot\left(\frac{\sin\frac x2}{\frac x2}\right)^2\right]\xrightarrow[x\to0^+]{}$$

$$\rightarrow1\left[1\cdot1-\frac22\cdot\frac14\cdot1\right]=1-\frac14=\frac34$$