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Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$

I need to solve for $a$ and $b$, so here we go,

$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right)$

$= \lim\limits_{x \to ∞} x \left(2 +(3+x) \left(\ln (1+\frac{a}{x}) - \ln(1+\frac{b}{x}) \right) \right)$

$=\lim\limits_{x \to ∞} x \left(2 +(3+x)\left( \dfrac{a-b}{x} \right) \right)$

$=\lim\limits_{x \to ∞} \left(2x +(3+x)\left( a-b \right) \right)$

$=\lim\limits_{x \to ∞} \left(2x + 3(a-b) + x(a-b) \right) $

Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$

But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.

What do I do now? And where exactly have I gone wrong?

Thank you!

William
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5 Answers5

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Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $A\neq B$ but such replacements always represent some hidden steps which the the author assumes to be obvious (an example of such a replacement is in my solution where the factor $x$ is replaced by $x+b$ and it is based on the fact that $x/(x+b) \to 1$ as $x\to\infty$). Also see this answer for details about such replacements.

The given limit condition implies that $(x+3)\log((x+a)/(x+b))\to - 2$ Then clearly $a\neq b$ and we have $$(a-b)\cdot\frac{x+3}{x+b}\cdot\dfrac{\log\left(1+\dfrac{a-b}{x+b}\right)}{\dfrac{a-b}{x+b}}\to - 2$$ Thus $a-b=-2$ or $a=b-2$. Using this value of $a$ in original limit we see that $$x\left(2+(x+3)\log\left(1-\frac{2}{x+b}\right)\right)\to 2$$ Replacing the first factor by $x+b$ and putting $x+b=2/t$ we see that $$\lim_{t\to 0^{+}}\frac{1}{t}\left(2+\left(\frac{2}{t}+3-b\right)\log(1-t)\right)=1$$ or $$\frac{2t+2\log(1-t)}{t^2}+(3-b)\frac{\log(1-t)}{t}\to 1$$ By L'Hospital's Rule or Taylor series the first fraction tends to $-1$ so that $-1+b-3=1$ or $b=5$ and $a=3$.

Paramanand Singh
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  • Thank you for such a detailed answer, I didn't want the solution like the most answers here, I only wanted someone to point out mistakes in mine, like you did. I knew those rules, I was almost certain I could seperate the log expression from the rest of the expression. Guess I was quick to jump to the conclusions. – William Jan 19 '19 at 09:39
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Hint: you have to expand $\log (1+y)$ up to the term in $y^{2}$ to answer this question. $\log (1+y)=y-\frac {y^{2}} 2+0(y^{2})$

Kavi Rama Murthy
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  • Then my next obvious question would be why doesn't $\lim\limits_{x\to 0} \dfrac{\ln ( 1+ x)}{x} = 1 $ work here? When $x \to \infty \Rightarrow \frac{a}{x} \to 0$ in which case the identity should work right? All I've done is multiply and divide by $\frac{a}{x}$ I know Maclaurin Series may work here but still why is this wrong? – William Jan 19 '19 at 05:52
  • @William: well the limit $\lim\limits _{x\to 0}\dfrac{\log (1+x)}{x}=1$ does work here (see my answer) but it does not mean that you can replace $\log(1+x)$ with $x$ but rather it means that you can replace the expression $\lim\limits _{x\to 0}\dfrac{\log(1+x)}{x}$ with $1$. See more details in beginning of my answer. – Paramanand Singh Jan 19 '19 at 08:51
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Hint

First of all, let $x=\frac 1y$ to make the expression $$A=\frac{(3 y+1) \log \left(\frac{1+a y}{1+b y}\right)+2 y}{y^2}$$ Now, using Taylor series, you should have $$A=\frac{a-b+2}{y}+\frac{1}{2} \left(-a^2+6 a+b^2-6 b\right)+\frac{1}{6} y \left(2 a^3-9 a^2-2 b^3+9 b^2\right)+O\left(y^2\right)$$ So $$a-b+2=0 \qquad \text{and} \qquad \frac{1}{2} \left(-a^2+6 a+b^2-6 b\right)=2$$ This is easy to solve.

Claude Leibovici
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Set $1/x=h$

$$\lim_{h\to0^+}\dfrac{2h+(3h+1)\{\ln(1+ah)-\ln(1+bh)\}}{h^2}$$

$$=3\left(a\lim_{h\to0^+}\dfrac{\ln(1+ah)-1}{ah}-b\lim_{h\to0^+}\dfrac{\ln(1+bh)-1}{bh}\right)+\lim_{h\to0^+}\dfrac{2h+\ln(1+ah)-\ln(1+bh)}{h^2}$$

$$=3(a-b)+\lim_{h\to0^+}\dfrac{2h+\ln(1+ah)-\ln(1+bh)}{h^2}$$

$$=3(a-b)+a^2\lim_{h\to0^+}\dfrac{\ln(1+ah)-ah}{(ah)^2}-b^2\lim_{h\to0^+}\dfrac{\ln(1+bh)-bh}{(bh)^2}+\lim_{h\to0^+}\dfrac{(2+a-b)h}{h^2}$$

$$=3(a-b)+\dfrac{b^2-a^2}2+\lim_{h\to0^+}\dfrac{(2+a-b)h}{h^2}$$ using Are all limits solvable without L'Hôpital Rule or Series Expansion,

Clearly, we need $2+a-b=0$ and $3(a-b)+\dfrac{b^2-a^2}2=2$

lab bhattacharjee
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Since the limit exists, $2x$ and $x(a−b)$ must cancel out so $a−b=2$.

You wanted to say $2+a-b=0 \Rightarrow \color{red}{a-b=-2}$.

And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+\cdots)$, which is multiplied by $x$. In other words, you can not apply the asymptotics to an addend, but must apply to a factor.

Alternatively: $$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2 \Rightarrow \\ \lim\limits_{x \to ∞} \frac{2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right)}{\frac 1x} = 2 \ \ \ (*)$$ For the limit to exist: $$\lim\limits_{x \to ∞} \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 0 \Rightarrow \cdots \Rightarrow a-b=-2.$$ Now use L'Hospital's rule for $(*)$ twice: $$\lim\limits_{x \to ∞} \frac{\ln (x+a) +\frac{3+x}{x+a}- \ln (x+b) -\frac{3+x}{x+b}}{-\frac 1{x^2}} = \\ =\lim\limits_{x \to ∞} \frac{\frac1{x+a}+\frac{a-3}{(x+a)^2}-\frac1{x+b}-\frac{b-3}{(x+b)^2}}{\frac 2{x^3}} = \\ =\lim_{x\to\infty} -\frac{x^3 (a - b) (2 a b + a x - 3 a + b x - 3 b - 6 x)}{2 (x+a)^2 (x+b)^2}=2 \Rightarrow \\ \color{red}{a+b-6=2}.$$ Hence: $$\begin{cases}a-b=-2 \\ a+b-6=2 \end{cases} \Rightarrow a=3, b=5.$$

farruhota
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