Limitar el problema con la quinta raíz

Question

PS

¿Cómo resolver este límite sin la regla de L'Hopital y las series de Taylor?

Answer 1

Tenemos$$\dfrac1{a-b} = \dfrac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^5-b^5}$ $ Tomando$a=\sqrt[5]{1+5x}$ y$b=1+x$, obtenemos que \begin{align} \dfrac1{\sqrt[5]{1+5x}-1-x} = \dfrac{(1+5x)^{4/5} + (1+5x)^{3/5}(1+x) + (1+5x)^{2/5}(1+x)^{2} + (1+5x)^{1/5}(1+x)^3 + (1+x)^4}{(1+5x)-(1+x)^5} \end {align} Por lo tanto, \begin{align} \dfrac{x^2}{\sqrt[5]{1+5x}-1-x} & = \dfrac{(1+5x)^{4/5} + (1+5x)^{3/5}(1+x) + (1+5x)^{2/5}(1+x)^{2} + (1+5x)^{1/5}(1+x)^3 + (1+x)^4}{-10-15x-5x^2-x^3} \end {align} Ahora tomando el límite como$x \to 0$, obtenemos el límite para ser$$L = \dfrac{1+1+1+1+1}{-10} = -\dfrac12$ $

Answer 2

Usemos la sustitución simple$1 + 5x = t^{5}$ para que$x = (t^{5} - 1)/5$. Luego, como$x \to 0$ tenemos$t \to 1$. Por lo tanto \begin{align} L &= \lim_{x \to 0}\frac{x^{2}}{\sqrt[5]{1 + 5x} - 1 - x}\notag\\ &= \lim_{t \to 1}\dfrac{\left(\dfrac{t^{5} - 1}{5}\right)^{2}}{t - 1 - \dfrac{t^{5} - 1}{5}}\notag\\ &= \frac{1}{5}\lim_{t \to 1}\frac{(t^{5} - 1)^{2}}{5(t - 1) - (t^{5} - 1)}\notag\\ &= \frac{1}{5}\lim_{t \to 1}\frac{(t - 1)^{2}}{5(t - 1) - (t^{5} - 1)}\cdot\left(\frac{t^{5} - 1}{t - 1}\right)^{2}\notag\\ &= \frac{1}{5}\lim_{t \to 1}\frac{t - 1}{5 - (t^{4} + t^{3} + t^{2} + t + 1)}\cdot\left(5\right)^{2}\notag\\ &= -5\lim_{t \to 1}\frac{t - 1}{(t^{4} + t^{3} + t^{2} + t) - 4}\notag\\ &= -5\lim_{t \to 1}\frac{t - 1}{(t^{4} - 1) + (t^{3} - 1) + (t^{2} - 1) + (t - 1)}\notag\\ &= -5\lim_{t \to 1}\dfrac{1}{\dfrac{t^{4} - 1}{t - 1} + \dfrac{t^{3} - 1}{t - 1} + \dfrac{t^{2} - 1}{t - 1} + 1}\notag\\ &= -5\cdot\frac{1}{4 + 3 + 2 + 1}\notag\\ &= -\frac{5}{10} = -\frac{1}{2}\notag \end {align}

Answer 3

$$\lim_{x\to 0}\frac{x^2}{\sqrt[5]{1+5x}-1-x}=\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\left(x^2\right)}{\frac{\text{d}}{\text{d}x}\left(\sqrt[5]{1+5x}-1-x\right)}=$ $$$\lim_{x\to 0}\frac{2x}{\frac{1}{\left(5x+1\right)^{\frac{4}{5}}}-1}=\lim_{x\to 0}-\frac{2x(5x+1)^{\frac{4}{5}}}{(5x+1)^{\frac{4}{5}}-1}=$ $$$-2\left(\lim_{x\to 0}(5x+1)^{\frac{4}{5}}\right)\left(\lim_{x\to 0}\frac{x}{(5x+1)^{\frac{4}{5}}-1}\right)=$ $$$-2\left((5\cdot 0+1)^{\frac{4}{5}}\right)\left(\lim_{x\to 0}\frac{x}{(5x+1)^{\frac{4}{5}}-1}\right)=$ $$$-2\left((0+1)^{\frac{4}{5}}\right)\left(\lim_{x\to 0}\frac{x}{(5x+1)^{\frac{4}{5}}-1}\right)=$ $$$-2\left((1)^{\frac{4}{5}}\right)\left(\lim_{x\to 0}\frac{x}{(5x+1)^{\frac{4}{5}}-1}\right)=$ $$$-2\left(1\right)\left(\lim_{x\to 0}\frac{x}{(5x+1)^{\frac{4}{5}}-1}\right)=$ $$$-2\left(\lim_{x\to 0}\frac{x}{(5x+1)^{\frac{4}{5}}-1}\right)=$ $$$-2\left(\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\left(x\right)}{\frac{\text{d}}{\text{d}x}\left((5x+1)^{\frac{4}{5}}-1\right)}\right)=$ $$$-2\left(\lim_{x\to 0}\frac{1}{\frac{4}{\sqrt[5]{5x+1}}}\right)=$ $$$-2\left(\lim_{x\to 0}\frac{\sqrt[5]{5x+1}}{4}\right)=$ $$$-2\left(\frac{\sqrt[5]{5\cdot 0+1}}{4}\right)=$ $$$-2\left(\frac{\sqrt[5]{1}}{4}\right)=$ $$$-2\left(\frac{1}{4}\right)=-\frac{1}{2}$ $