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I want to solve the following limit problem $$\lim_{x \rightarrow \infty} \bigg[ (x + a)\log \big( \frac{x+a}{x+b} \big) \bigg] $$

A small simulation with a = 5 and b = 2 leads to a result of 3 which I think is correct and the right answer should be (a - b), somehow I am unable to prove it.

Here is the code to reproduce the result

def f(alpha):
    return (5 + alpha) * np.log((5 + alpha) / (2 + alpha))


alpha_list = np.linspace(10,1000)
plt.plot(alpha,[f(a) for a in alpha_list])

yields the following plot,

enter image description here

Thanks in advance.

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5 Answers 5

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Use the limit definition of $e$

$$-(x+a)\log\left(\frac{x+b}{x+a}\right) = -\log\left(1+\frac{b-a}{x+a}\right)^{x+a}$$ $$\longrightarrow -\log e^{b-a} = a-b$$

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  • $\begingroup$ Elegant solution. $\endgroup$ Commented Apr 12, 2021 at 20:02
  • $\begingroup$ Really liked the solution!! thanks $\endgroup$ Commented Apr 12, 2021 at 20:04
  • $\begingroup$ Thanks for the kind words everyone! $\endgroup$ Commented Apr 12, 2021 at 21:00
  • $\begingroup$ (+1) great way of thinking about it! $\endgroup$ Commented Apr 13, 2021 at 10:16
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Lets think logically about what is going on here, as $x\to\infty$: $$\log\left(\frac{x+a}{x+b}\right)\approx\log\left(\frac xx\right)=\log(1)=0$$ as $x$ will be much larger than $a,b$. Now in the second part: $$x+a\to\infty$$ and this is pretty obvious. So, to put it crudely we have a $0\times\infty$ situation. If you want to use L'Hopital's rule you can do the following: $$(x+a)\log\left(\frac{x+a}{x+b}\right)=\frac{\log(x+a)-\log(x+b)}{\frac{1}{x+a}}$$ Now both the top and bottom of this fraction tend to zero, so we can say: $$\lim_{x\to\infty^+}\frac{\log(x+a)-\log(x+b)}{\frac{1}{x+a}}=\lim_{x\to\infty^+}\frac{\frac{1}{x+a}-\frac{1}{x+b}}{-\frac{1}{(x+a)^2}}$$ $$=\lim_{x\to\infty^+}\frac{(x+a)^2}{(x+b)}-(x+a)$$ which we can rewrite as: $$\lim_{x\to\infty^+}\frac{(x+a)\left[(x+a)-(x+b)\right]}{(x+b)}=\lim_{x\to\infty^+}\frac{(x+a)(a-b)}{(x+b)}=\lim_{x\to\infty^+}(a-b)=a-b$$ which is probably the easiest way to do it

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With $ y=\frac 1x $, it becomes

$$\lim_{y\to 0^+}\frac{1+ay}{y}\Bigl(\ln(1+ay)-\ln(1+by)\Bigr)=$$

$$\lim_{y\to 0^+}(1+ay)\Bigl(\frac{\ln(1+ay)}{ay}a-\frac{\ln(1+by)}{by}b\Bigr)=$$ $$1.(1.a-1.b)=a-b$$

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Note that $$ \begin{align} (x+a)\log\left(\frac{x+a}{x+b}\right) &=(x+a)\int_{x+b}^{x+a}\frac1t\,\mathrm{d}t\tag{1a}\\ &=(x+a)\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: $\log(t)$ is the integral of $1/t$
$\text{(1b)}$: substitute $t\mapsto xt$

Next, we have the bounds $$ \overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+a/x}}^{\substack{\text{minimum of}\\\text{integrand}}}\le\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\le\overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+b/x}}^{\substack{\text{maximum of}\\\text{integrand}}}\tag2 $$ Multiplying $(2)$ by $x+a$ and then applying $(1)$, we get $$ a-b\le(x+a)\log\left(\frac{x+a}{x+b}\right)\le(a-b)\frac{x+a}{x+b}\tag3 $$ Graphically, we can see the function squeezed between two functions whose limits are easier to compute:

enter image description here

The Squeeze Theorem then gives $$ \lim_{x\to\infty}(x+a)\log\left(\frac{x+a}{x+b}\right)=a-b\tag4 $$

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  • $\begingroup$ Very interesting, just wondering, is squeeze theorem true for all (real-valued) functions? $\endgroup$ Commented Apr 13, 2021 at 13:44
  • $\begingroup$ If $f_-(x)\le f(x)\le f_+(x)$ for all $x$ near $a$ and $\lim\limits_{x\to a}f_-(x)=\lim\limits_{x\to a}f_+(x)=L$, then $\lim\limits_{x\to a}f(x)=L$, where "near" $\infty$ means sufficiently large. $\endgroup$ Commented Apr 13, 2021 at 15:04
  • $\begingroup$ Note that $\lim\limits_{x\to\infty}(a-b)=\lim\limits_{x\to\infty}\frac{x+a}{x+b}(a-b)=a-b$ $\endgroup$ Commented Apr 13, 2021 at 15:08
  • $\begingroup$ I have added a link to the Squeeze Theorem. $\endgroup$ Commented Apr 13, 2021 at 15:16
  • $\begingroup$ I have also added an illustration showing the "squeezing". $\endgroup$ Commented Apr 13, 2021 at 22:55
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A straightforward solution uses the Taylor series for $\log(1+x)$.

$$\log\frac{x+a}{x+b}=\log\left(1+\frac{a-b}{x+b}\right)=\frac{a-b}{x+b}+O\left(\frac1{(x+b)^2}\right)$$ $$\lim_{x\to\infty}(x+a)\log\frac{x+a}{x+b}=\lim_{x\to\infty}(a-b)\frac{x+a}{x+b}=a-b$$

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  • $\begingroup$ Your last step is l'hopitals? $\endgroup$ Commented Apr 12, 2021 at 20:14
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    $\begingroup$ @HirakSarkar If you mean how to get $\lim_{x\to\infty}\frac{x+a}{x+b}=1$, you can use L'Hospital, or just divide numerator and denominator by $x$. $\endgroup$ Commented Apr 12, 2021 at 20:16

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