How to evaluate $\lim _{x\to \infty }\left(x^3\ln\left(\frac{x}{x-1}\right)-x\sqrt{x^2+x}\right)$?

Question

$$\lim _{x\to \infty }\left(x^3\ln\left(\frac{x}{x-1}\right)-x\sqrt{x^2+x}\right)$$ 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)?

Answer 1

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

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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.)

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