Limit problem with fifth root

Question

$$\lim_{x\to 0}\frac{x^2}{\sqrt[5]{1+5x}-1-x}$$

How to solve this limit without L'Hopital rule and Taylor series?

Answer 1

We have $$\dfrac1{a-b} = \dfrac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^5-b^5}$$ Taking $a=\sqrt[5]{1+5x}$ and $b=1+x$, we obtain that \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} Hence, \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} Now taking the limit as $x \to 0$, we obtain the limit to be $$L = \dfrac{1+1+1+1+1}{-10} = -\dfrac12$$

Answer 2

Let's use the simple substitution $1 + 5x = t^{5}$ so that $x = (t^{5} - 1)/5$. Then as $x \to 0$ we have $t \to 1$. Therefore \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}