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This equation goes like this: $$\lim_{x \to \infty}\ln\left(\frac{x}{x+1}\right) = 0$$

Can you help me understand the logic behind the above equation?

Lucas
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    Because $\frac{x}{x+1}\rightarrow 1$ and $\ln$ is continuous. – Gregory Grant Jul 18 '17 at 16:20
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    $\lim \limits_{x \to \infty}\ln\left(\frac{x}{x+1}\right) = \ln\left(\lim\limits_{x \to \infty}\frac{x}{x+1}\right)$ as $\ln$ is continuous. https://math.stackexchange.com/questions/1518738/when-can-i-move-the-limit-operand-into-a-function – Shuri2060 Jul 18 '17 at 16:23

5 Answers5

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The logarithm function is continuous (so that the limit of the logarithm is the logarithm of the limit) and its argument tends to $1$ (because the inverse, $1+1/x$, tends to $1$).

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$$\lim_{x\to\infty}\dfrac{x}{x+1}=1$$ $$\ln(1)=0$$

Riccardo.Alestra
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substitute $u\mapsto \frac{x}{x+1}$ now you're left with

$$ \lim_{u\to 1} \ln u = \ln 1 = 0 $$

In English: As $x$ gets bigger, the fraction $\frac{x}{x+1}$ gets closer and closer to $1$. Write out several terms to see this: $\frac 3 4$, $\frac 4 5$, $\ldots$, $\frac{999}{1000}$, $\ldots$, $\frac{999\ 999}{1\ 000\ 000} \to 1$. And we know $\ln(1) = 0$.

Dando18
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See the Limits of compositions of functions theorem

$\left\lbrace\begin{array}l \lim \limits_{x\to +\infty} \dfrac{x}{x+1} =1 \\\\ \lim \limits_{X\to 1} \ln X =0 \end{array}\right.\implies \lim \limits_{x\to +\infty} \ln\left(\dfrac{x}{x+1}\right) =0$

Stu
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$$\lim_{x \to \infty}\ln \left(\frac{x}{x+1}\right)=\ln \left(\lim_{x \to \infty}\frac{x}{x+1}\right)=\ln \left(\lim_{x \to \infty}\frac{1}{1+\frac1x}\right)=\ln 1 = 0$$

The first equality is due to continuity of $\ln$ function.

Siong Thye Goh
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