Solving the limit $\lim_{x\rightarrow+\infty}(\sqrt[5]{x^5-x^4}-x)$

Question

Recently, I am struggling to solve the limit: $$\lim_{x\rightarrow+\infty}(\sqrt[5]{x^5-x^4}-x)$$ If I try to make some fraction with nominator $-x^4$ and some irrational denominator by multiplying, it becomes more complex. Can anyone help about this with more easier way?

Answer 1

Hint:

Let $1/x=h\implies h\to0$

$$\sqrt[5]{x^5-x^4}=\sqrt[5]{\dfrac{1-h}{h^5}}=\dfrac{\sqrt[5]{1-h}}h$$

Now set $\sqrt[5]{1-h}=y\implies h=1-y^5$ to have $$-\lim_{y\to1}\dfrac{y-1}{y^5-1}=?$$

Answer 2

$$(\sqrt[5]{x^5-x^4}-x)=\frac{\sqrt[5]{1-x^{-1}}-1}{1/x}$$ $\sqrt[5]{1-x^{-1}}-1\rightarrow0$ and $1/x\rightarrow0$ as $x\rightarrow\infty$. So apply l'Hospital's rule.

Answer 3

Let $a=\sqrt[5]{x^5-x^4}$ and $b=x$. Then\begin{align}\sqrt[5]{x^5-x^4}-x&=a-b\\&=\frac{(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)}{a^4+a^3b+a^2b^2+ab^3+b^4}\\&=\frac{a^5-b^5}{a^4+a^3b+a^2b^2+ab^3+b^4}\\&=\frac{-x^4}{\sqrt[5]{x^5-x^4}^4+\sqrt[5]{x^5-x^4}^3\,x+\sqrt[5]{x^5-x^4}^2\,x^2+\sqrt[5]{x^5-x^4}\,x^3+x^4}\\&=\frac{-1}{\left(\frac{\sqrt[5]{x^5-x^4}}x\right)^4+\left(\frac{\sqrt[5]{x^5-x^4}}x\right)^3+\left(\frac{\sqrt[5]{x^5-x^4}}x\right)^2+\frac{\sqrt[5]{x^5-x^4}}x+1}\\&=\frac{-1}{\left(\sqrt[5]{1-\frac1x}\,\right)^4+\left(\sqrt[5]{1-\frac1x}\,\right)^3+\left(\sqrt[5]{1-\frac1x}\,\right)^2+\sqrt[5]{1-\frac1x}+1}\end{align}and therefore\begin{align}\lim_{x\to+\infty}\sqrt[5]{x^5-x^4}-x&=\lim_{x\to+\infty}\frac{-1}{\left(\sqrt[5]{1-\frac1x}\,\right)^4+\left(\sqrt[5]{1-\frac1x}\,\right)^3+\left(\sqrt[5]{1-\frac1x}\,\right)^2+\sqrt[5]{1-\frac1x}+1}\\&=-\frac15.\end{align}

Answer 4

$$\lim_{x\to\infty}(\sqrt[5]{x^5-x^4}-x)=\lim_{x\to\infty}(x\sqrt[5]{1-\frac{1}{x}}-x)$$ By Applying Binomial Theorem ($a<<1 \implies (1-a)^x = (1-xa$)) $$\lim_{x\to\infty}(x(1-\frac{1}{5x})-x)=\lim_{x\to\infty}((x-\frac{1}{5})-x)=\lim_{x\to\infty}(x-\frac{1}{5}-x)=-\frac{1}{5}$$

Answer 5

A slight twist :

Using bhatthachargee's substitution $1/x =:h$ , $h \rightarrow 0$, leads to the expression:

$ f(h):=- \dfrac{\sqrt[5]{1-h} -1}{-h}$.

Note:

$\lim_{h \rightarrow 0}f(h) = $

$-(\dfrac{d}{dx} \sqrt[5]{x})_{x=1}= -(1/5)$

Answer 6

By definition of derivative at $t=0$. set $t=\frac1x$

$$\lim_{x\rightarrow+\infty}(\sqrt[5]{x^5-x^4}-x) =\lim_{t\to 0}\frac{\sqrt[5]{1-t}-1}{t-0} =(\sqrt[5]{1-t})'|_{t=0}=-\frac15 $$