calcular el límite (no l'Hospital de la regla)

Question

Necesito calcular $$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}.$$ No puedo usar la regla de l'Hospital de.

Answer 1

Como lcm $(2,3)=6, $ deje $(\cos x)^\frac16=y$

$\displaystyle\implies (\cos x)^\frac12=y^3,(\cos x)^\frac13=y^2$ $\displaystyle \cos x=y^6\implies\sin^2x=1-y^{12}$

$$\lim_{x\to0}\frac{(\cos x)^\frac12-(\cos x)^\frac13}{\sin^2x}=\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}\frac{-y^2(1-y)}{(1-y)(1+y+\cdots+y^{10}+y^{11})}$$

Podemos cancelar $1-y$ $1-y\ne0$ $y\to1$

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Alternativamente, $$\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}(-y^2)\cdot\frac1{\lim_{y\to1}\frac{y^{12}-1}{y-1}}$$

De nuevo, $$\lim_{y\to1}\frac{y^{12}-1}{y-1}=\lim_{y\to1}\frac{y^{12}-1^{12}}{y-1}=\frac{d(y^{12})}{dy}_{(\text{ at }y=1)}$$

Answer 2

$$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{\cos x} -1+1- \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{1+\cos x-1} -1- (\sqrt[3]{1+\cos x-1}-1)}{\cos x-1}\cdot\frac{\cos x-1}{\sin^2x}=\lim_{x\to 0}[\frac{(1+\cos x-1)^\frac{1}{2}-1}{\cos x-1}-\frac{(1+\cos x-1)^\frac{1}{3}-1}{\cos x-1}]\cdot\frac{-1}{2}=(\frac{1}{2}-\frac{1}{3})\cdot\frac{-1}{2}=-\frac{1}{12}$$

Hemos aplicado:$$ \lim_{t\to 0}\frac{(1+t)^r-1}{t} =r$$