Finding the limit of $f(x)$ tends to infinity How do you find this limit?
$$\lim_{x \rightarrow \infty} \sqrt[5]{x^5-x^4} -x$$
I was given a clue to use L'Hospital's rule.
I did it this way:
UPDATE 1: 
$$
\begin{align*}
\lim_{x \rightarrow \infty} \sqrt[5]{x^5-x^4} -x 
&= \lim_{x \rightarrow \infty} x\begin{pmatrix}\sqrt[5]{1-\frac 1 x} -1\end{pmatrix}\\
&= \lim_{x \rightarrow \infty} \frac{\sqrt[5]{1-\frac 1 x} -1}{\frac1x}
\end{align*}
$$
Applying L' Hospital's,
$$
\begin{align*}
\lim_{x \rightarrow \infty} \frac{\sqrt[5]{1-\frac 1 x} -1}{\frac1x}&=
\lim_{x \rightarrow \infty} \frac{0.2\begin{pmatrix}1-\frac 1 x\end{pmatrix}^{-0.8}\begin{pmatrix}-x^{-2}\end{pmatrix}(-1)} {\begin{pmatrix}-x^{-2}\end{pmatrix}}\\
&= -0.2
\end{align*}
$$
However the answer is $0.2$, so I would like to clarify the correct use of L'Hospital's
 A: You got the answer, but I'd like to note something different. I see you are doing derivations, so I am writing an answer based on it. We say the function $\alpha(x)$ is very small at $x\to a$ when $$\lim\alpha(x)\to 0$$ We can prove that by using Taylor expansion that $\sqrt[n]{1+\alpha(x)}-1\sim\frac{\alpha(x)}{n}$. So $$\frac{\sqrt[5]{1-k}-1}k~\sim~\frac{-k/5}{k}=-1/5$$ when $k\to 0$.
A: Your working out is fine and you've shown all the steps now.  $$\sqrt[5]{x^5 - x^4} < x$$ and so a negative limit is more likely than a positive limit :)
A: If we are not compelled to use L'Hospital's Rule,
$$\lim_{x \rightarrow \infty} \sqrt[5]{x^5-x^4} -x$$
$$=\lim_{y\to0}\frac{(1-y)^\frac15-1}y$$
$$=\lim_{y\to0}\frac{(1-y)-1}{y\{(1-y)^\frac45+(1-y)^\frac35+(1-y)^\frac25+(1-y)^\frac15+1\}}$$ as $ a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$
$$=\frac{-1}{1+1+1+1+1}\text {  as } y\to0\implies y\ne0$$
A: Yes, you are using L'Hopital's rule correctly, and the answer is $-\frac{1}{5}$. The steps are correct, and I double-checked the answer: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427epe5q2eamff
Not a big deal, but please note 'L'Hopital' does not have an 's' in it.
