$$\lim _{x\to \infty }\left(x^3\ln\left(\frac{x}{x-1}\right)-x\sqrt{x^2+x}\right)$$ I tried like that: $$\lim _{x\to \infty }\left(x^3\left(\frac{1}{x-1}\right)-x^2\sqrt{1+\frac{1}{x}}\right)=\lim _{x\to \infty }\left(\left(\frac{x^3}{x-1}\right)-x^2\right)$$ $$=\lim _{x\to \infty }\left(\frac{x^2}{x-1}\right)=\infty$$ But the result is wrong, actually it has to come $11/24$ Can you please show me how to solve it (without l'Hospital)?


1 Answer 1


About what you did: In your attempt, where did the $\ln$ go?

An approach: Here is a solution using Taylor expansions: we will use the fact that, when $u\to 0$, $$\begin{align} -\ln(1-u) &= u + \frac{u^2}{2} + \frac{u^3}{3} + o(u^3) \\ \sqrt{1+u} &= 1 + \frac{u}{2} - \frac{u^2}{8} + o(u^2) \end{align}$$ and the fact that $\frac{1}{x} \xrightarrow[x\to\infty]{} 0$ (this will be "our $u$").

(Intuitively, we go to order $3$ in the first and only order $2$ in the second since, for our application, we will multiply by ${x^3}=u^{-3}$ and $x^2=u^{-2}$, respectively: we want to do an expansion of both terms of the difference, up until we get a constant.)

  • First term: $$x^3\ln \frac{x}{x-1} = -x^3 \ln \left(1-\frac{1}{x}\right) = x^3 \left(\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{3x^3}+o\left(\frac{1}{x^3}\right)\right) = x^2 + \frac{x}{2} + \frac{1}{3} + o(1)$$

  • Second term: $$ x\sqrt{x^2+x} = x^2\sqrt{1+\frac{1}{x}} = x^2\left(1+\frac{1}{2x}-\frac{1}{8x^2}+o\left(\frac{1}{x^3}\right)\right) = x^2 + \frac{x}{2} - \frac{1}{8} + o(1)$$

  • Now, the difference of the two: $$x^3\ln \frac{x}{x-1} - x\sqrt{x^2+x} = \frac{1}{3} + \frac{1}{8} + o(1) = \frac{11}{24} + o(1) \xrightarrow[x\to\infty]{} \frac{11}{24}. $$

  • $\begingroup$ I know that if the argument of the $ln$ tend to $1$ i can delete the ln and add $-1$, is it right? $\endgroup$
    – Amarildo
    Jan 28, 2016 at 21:41
  • $\begingroup$ That's not enough. What you describe is basically a Taylor expansion up to order $1$ ($\ln(1+u) = u + o(u)$; you also did $\sqrt{1+u} = 1 + o(1)$), but here you actually need order $3$: the terms you remove/neglect are multiplied by $x^3$, so even something as small as $1/x^2$ cannot be neglected that way: multiplied by $x^3$, it becomes $x$, and you are considering as negligible something that goes to infinity. (And something, actually, that would conveniently cancel with other terms of the square root, that you also threw away as negligible while they really were not) $\endgroup$
    – Clement C.
    Jan 28, 2016 at 21:44
  • $\begingroup$ I understand, thank you very much :) $\endgroup$
    – Amarildo
    Jan 28, 2016 at 21:46
  • $\begingroup$ You're welcome! $\endgroup$
    – Clement C.
    Jan 28, 2016 at 21:47
  • $\begingroup$ Concise and complete! Nicely presented. +1 - Mark $\endgroup$
    – Mark Viola
    Jan 29, 2016 at 0:14

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