# 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

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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 $k\to 0$.

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Yes indeed, nice answer! +1 –  amWhy Mar 1 '13 at 0:55

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 :)

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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 $a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$

$$=\frac{-1}{1+1+1+1+1}\text { as } y\to0\implies y\ne0$$

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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.

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Hi, please check my updated solution? –  bryansis2010 Feb 28 '13 at 14:29
L'Hospital did indeed spell his name that way. Later on, French orthography changed so that the "os" became "ô". (One sees this in other words like "hôtel", originally "hostel".) So either the original spelling, "L'Hospital", or the modern spelling, "L'Hôpital", is correct. In any case "L'Hopital", with no circumflex, is incorrect. –  MJD Feb 28 '13 at 14:47
Oh I agree, please spell it with a circumflex. I'm just lazy typing on the computer. I'm happy as long as you don't pronounce it "hospital" like the place where you go when you're sick :) –  ferson2020 Feb 28 '13 at 14:53
I indeed pronounced it as "hospital" initially...but then Wikipedia mentioned to pronounce it with just the silent "s", so yes, Hopital is the pronunciation but "Hospital" can be the spelling. –  bryansis2010 Feb 28 '13 at 15:02