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I have been working around with limits involving infinity and was stuck with this limit: $$\lim_{x\to{\infty}}[(x^5+10x^4+3)^c - x]$$ Can this limit have some finite value for any real value of c? If yes then what is c and the limit? I tried putting x=1/t and letting $$\lim_{t\to 0}$$Then I used L'Hospital's Rule but ended up having another 0/0 form. Please help me.

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  • $\begingroup$ It will turn out that $c=1/5$. $\endgroup$ – André Nicolas Oct 2 '15 at 6:04
  • $\begingroup$ Can you explain how? $\endgroup$ – Piyush Oct 2 '15 at 6:05
  • $\begingroup$ Because $(x^5-10x^4+3)^{1/5}=x+.....$ $\endgroup$ – Ahmed S. Attaalla Oct 2 '15 at 6:06
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    $\begingroup$ Maybe for a detailed proof multiply top and bottom by $u^4+u^3v+u^2v^2+uv^3+v^4$ where $u=(x^5-10x^4+3)^{1/5}$ and $v=x$. And it should not be hard to show that the thing blows up if $c\gt 1/5$ and goes to $-\infty$ if $c\lt 1/5$. $\endgroup$ – André Nicolas Oct 2 '15 at 6:12
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If $c \leq 0$, then $(x^5 + 10 x^4 + \cdots)^c$ is bounded as $x \to \infty$, and therefore the quantity $(x^5 + 10 x^4 + \cdots)^c - x$ in the limit is unbounded. So, for any $c$ for which the limit is finite, we have $c > 0$.

Now, expanding $(x^5 + 10 x^4 + \cdots)^c$ in a series about $x = \infty$ gives $$(x^5 + 10 x^4 + \cdots)^c = x^{5 c} + 10 c x^{5 c - 1} + R_c(x),$$ where $R_c(x)$ is a remainder term (which depends on $c$) of order $O(x^{5 c - 2})$. We can write a series expansion of the expression in the limit, namely: $$(x^5 + 10 x^4 + \cdots)^c - x = - x + x^{5 c} + 10 c x^{5 c - 1} + R_c(x).$$

Now, if $c < \frac{1}{5}$, the first term dominates all of the others as $x \to \infty$, in which case the limit is $\lim_{x \to \infty} -x = -\infty$. On the other hand, if $c > \frac{1}{5}$, the second term dominates all the others, and the limit is $\lim_{x \to \infty} x^{5c} = \infty$. In the remaining case, $c = \frac{1}{5}$, we see that the first two terms cancel, and substituting gives that the series has the form $$2 + R_{1/5}(x),$$ and this remainder term has order $O(x^{-1})$. In particular, $\lim_{x \to \infty} R_{1/5}(x) = 0$, and hence $$\boxed{\lim_{x \to \infty} [(x^5 + 10 x^4 + \cdots)^c - x] = 2} .$$

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$$\lim_{x\to{\infty}}[(x^5+10x^4+3)^c - x]$$ Let's go off of the assumption that $c=\frac{1}{5}$ (noted in the comments... can be noticed by observing the highest power of $x$ in the polynomial. We want to have the root be the inverse of this in order to have the highest power of $x$ be $1$ and cancel with the other $x$) $$\lim_{x\to{\infty}}[(x^5+10x^4+3)^\frac{1}{5} - x]$$ $$=\lim_{x\to{\infty}}[x+2-\frac{8}{x}+\frac{48}{x^2}-\frac{336}{x^3}+\frac{12771}{5 x^4}...-x]$$ $$= \color{red}{2}$$ Note that this is called a Laurent series (they differ from Taylor series in that they can have negative powers of $x$, which we needed here)

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  • $\begingroup$ A more sensible way of doing this is to turn the «$\infty-\infty$» indeterminacy into a «$\infty/\infty$» and then use L'Hôpital's rule., as explained here, for example. $\endgroup$ – Mariano Suárez-Álvarez Oct 2 '15 at 6:23
  • $\begingroup$ @MarianoSuárez-Alvarez True, but it's much faster to calculate a Laurent series using a math software then to do the substitutions (however, I'll agree that knowing the subs is a good thing in the event you lack a computer or you wish to understand the principle) $\endgroup$ – Brevan Ellefsen Oct 2 '15 at 6:31
  • $\begingroup$ If you have math software available, then the fast thing is to simply ask it to compute the limit! :-) $\endgroup$ – Mariano Suárez-Álvarez Oct 2 '15 at 6:35
  • $\begingroup$ @MarianoSuárez-Alvarez touché XD Math software generally won't show you how it gets it answers though (WA will in some cases), so I figured it would be clearer if I showed the series rather than skipping to the solution. You're right though, I could probably expand my answer to describe how one gets the Laurent series, but I was unsure of the OP's math level and it intuitively makes sense that $(x^n + x^{n-1} +...+x+1)^\frac{1}{n} = x+....$ $\endgroup$ – Brevan Ellefsen Oct 2 '15 at 6:43
  • $\begingroup$ @BrevanEllefsen Could you update your answer to show that how to get the required Laurent series. $\endgroup$ – Piyush Oct 2 '15 at 9:47
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$$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$ Now substituting $a=(x^5+10x^4+3)^{1/5}$ , $b=x$ and taking limits we can show that corresponding function has a limit at infinity.

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