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I am trying to calculate this limit: $$\lim_{x \to \infty} x^2(\ln x-\ln (x-1))-x$$ The answer is $1/2$ but I am trying to verify this through proper means. I have tried L'Hospital's Rule by factoring out an $x$ and putting that as $\frac{1}{x}$ in the denominator (indeterminate form) but it becomes hopeless afterwards. Also I am a little hesitant about series involving the natural log because of restricted interval of convergence as $x$ is going to infinity. Is there a different approach how to do evaluate this limit? Thanks.

Zain Patel
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imranfat
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5 Answers5

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\begin{align*} \lim_{x \to \infty} [ x^2(\ln x-\ln (x-1))-x]&=\lim_{x \to \infty} \frac{\displaystyle\ln x-\ln (x-1)-\frac{1}{x}}{\displaystyle\frac{1}{x^2}}\\ &=\lim_{x \to \infty} \frac{\displaystyle\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x^2}}{\displaystyle\frac{-2}{x^3}}\quad (\text{ L'Hopital Rule})\\ &=\lim_{x \to \infty} \frac{\displaystyle\frac{-1}{x^2(x-1)}}{\displaystyle\frac{-2}{x^3}}\\ &=\frac{1}{2} \end{align*}

CY Aries
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  • Ah, instead of putting $1/x$ in the denominator, you put $1/x^2$ in the denominator so that the $ln$ terms have no longer an $x$ upfront and thus, their derivatives become rational functions!. Excellent and thank you! – imranfat May 15 '17 at 17:56
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HINT:

Set $1/x=h$ to get $$-\lim_{h\to0^+}\dfrac{\ln(1-h)+h}{h^2}$$

Now either use Taylor Expansion or L'Hospital

or See the third Question of Are all limits solvable without L'Hôpital Rule or Series Expansion

lab bhattacharjee
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$\begin{array}\\ \lim_{x \to \infty}x^2(\ln x-\ln (x-1))-x &= \lim_{x \to \infty} (-x^2(\ln (1-1/x))-x)\\ &= \lim_{y \to 0} (-\frac1{y^2}(\ln (1-y))-\frac1{y})\\ &= \lim_{y \to 0} (-\frac1{y^2}((y+\frac{y^2}{2}+O(\frac1{y^3}))-\frac1{y})\\ &= \lim_{y \to 0} ((\frac1{y}+\frac{1}{2}+O(\frac1{y}))-\frac1{y})\\ &= \lim_{y \to 0} (\frac{1}{2}+O(\frac1{y}))\\ &=\frac12\\ \end{array} $

You could also write $(-\frac1{y^2}(\ln (1-y))-\frac1{y}) =-\frac{\ln (1-y)+y}{y^2} $ and use L'Hopital.

marty cohen
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Using fact that $$\ln (1+u)=u-\frac {u^2}{2}(1+\epsilon (u)) $$ with $\lim_0 \epsilon (u)=0$, we get $$x^2 \ln (\frac{x}{x-1})-x=$$

$$x^2\ln(1+\frac {1}{x-1})-x =$$ $$=x^2 (\frac {1}{x-1}-\frac{1}{2 (x-1)^2} (1+\epsilon (x)) -x$$ with $\lim_{+\infty}\epsilon (x)=0$, $$=\frac {x}{x-1}-\frac {x^2}{2 (x-1)^2}\left(1+\epsilon (x)\right) $$

the limit is $$1-\frac {1}{2}=\frac 12$$

hamam_Abdallah
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$$\ln x - \ln(x-1)=\ln\frac{1}{1-\frac{1}{x}}=\frac{1}{x}+\frac{1}{2}\frac{1}{x^2}+\frac{1}{3}\frac{1}{x^3}+\cdots, $$ so $$x^2(\ln x - \ln(x-1))-x=\frac{1}{2}+\frac{1}{3}\frac{1}{x}+\cdots$$ and thus $$\lim_{x\rightarrow\infty}[x^2(\ln x - \ln(x-1))-x]=\frac{1}{2}.$$