Evaluating $\lim_{x\to-\infty}\ln(1+\frac{1}{x})-\frac{1}{x+1}$ with sign information

Question

Let $g(x)$= $\ln(1+\frac{1}{x})-\frac{1}{x+1}$

Clearly, $\lim\limits_{x\to+\infty} g(x) = 0^{+}$

I am having trouble wrapping my head around $\lim\limits_{x\to-\infty} g(x) $

We know for $|\frac{1}{x}|<1$ (which is true in our case)

$\ln(1+\frac{1}{x})=\frac{1}{x}-\frac{1}{2x^{2}}+\frac{1}{3x^{3}}-....$

and $\frac{1}{x+1} \sim \frac{1}{x}$ for $x\to-\infty$

so $g(x)= -\frac{1}{2x^{2}}+\frac{1}{3x^{3}}-....$ and hence $g(x)$ should tend to $0^{-}$ as $x\to-\infty$

but $g(x)$ tends to $0^{+}$ as $x\to-\infty$

What am I missing?

Edit- I think although $\frac{1}{x} \sim \frac{1}{x+1}$ as $x\to-\infty$ is true but since the other terms contain higher powers of $x$ in the denominator, it is probably not the best thing to do.

Answer 1

For any $\,x\in(-\infty,-1)\cup(1,+\infty)\,$ it results that

$\ln\left(1\!+\!\dfrac1x\right)=\dfrac1x-\dfrac1{2x^2}+o\!\left(\!\dfrac1{x^2}\!\right)\;\;,$

$\ln\left(1\!+\!\dfrac1x\right)-\dfrac1{x\!+\!1}=\dfrac1x-\dfrac1{x\!+\!1}-\dfrac1{2x^2}+o\!\left(\!\dfrac1{x^2}\!\right)\;\;,$

$\ln\left(1\!+\!\dfrac1x\right)-\dfrac1{x\!+\!1}=\dfrac{2x(x\!+\!1)-2x^2-(x\!+\!1)}{2x^2(x\!+\!1)}+o\!\left(\!\dfrac1{x^2}\!\right)\;\;,$

$\ln\left(1\!+\!\dfrac1x\right)-\dfrac1{x\!+\!1}=\dfrac{x\!-\!1}{2x^2(x\!+\!1)}+o\!\left(\!\dfrac1{x^2}\!\right)\;\;,$

$\ln\left(1\!+\!\dfrac1x\right)-\dfrac1{x\!+\!1}=\dfrac1{x^2}\left[\dfrac{x-1}{2(x\!+\!1)}+\dfrac{o\!\left(\!\frac1{x^2}\!\right)}{\frac1{x^2}}\right].$

Consequently ,

$\begin{align*}\lim\limits_{x\to-\infty}\left[\ln\left(\!1\!+\!\dfrac1x\!\right)\!-\!\dfrac1{x\!+\!1}\right]&=\!\!\lim\limits_{x\to-\infty}\dfrac1{x^2}\left[\dfrac{x\!-\!1}{2(x\!+\!1)}\!+\!\dfrac{o\!\left(\!\frac1{x^2}\!\right)}{\frac1{x^2}}\right]\!=\\[3pt]&=0^+\!\cdot\!\left[\dfrac12+0\right]=0^+\!\cdot\dfrac12=\\[3pt]&=0^+.\end{align*}$

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Addendum :

This addendum is similar to Greg Martin’s answer.

We could use the known inequality

$\ln(1+t)<t\quad$ for any $\,t>-1\,.\quad\color{blue}{(1)}$

For any $\,x\in(-\infty,-1)\cup(0,+\infty)\,$ it results that $\,t=-\dfrac1{x+1}>-1\,$ and, by applying $\,(1)\,,\,$ we get that

$\ln\left(1-\dfrac1{x+1}\right)<-\dfrac1{x+1}\;\;,\;\;$ consequently ,

$\ln\left(1+\dfrac1x\right)=-\ln\left(1-\dfrac1{x+1}\right)>\dfrac1{x+1}\,.$

Hence, your function

$g(x)=\ln\left(1+\dfrac1x\right)-\dfrac1{x+1}$

is positive on $\,(-\infty,-1)\cup(0,+\infty)\,.$

For that reason the limit of $\,g(x)\,$ as $\,x\to-\infty\,$ cannot be $\,0^-.$

Answer 2

No MacLaurin series are used, just the first derivative.

The function $f(x)=\log(1+{1\over x})-{1\over 1+x}$ satisfies $$f'(x)={1\over 1+x}-{1\over x}+{1\over (1+x)^2}\\ =-{1\over x(1+x)}+{1\over (1+x)^2} = -{1\over x(1+x)^2}$$ Therefore $f'(x)>0$ for $x<-1,$ i.e. $f(x)$ is increasing in $(-\infty,-1).$ As the limit at $-\infty$ is equal $0$ we get $$\lim_{x\to-\infty}f(x)=0^+$$

Remark The function $f(x)$ is increasing in $(-1,0)$ and decreasing in $(0,\infty).$ Thus $\lim_{x\to\infty}f(x)=0^+.$

Answer 3

It seems that the OP knows that $\lim_{x\to-\infty} g(x) = 0$ and is asking whether $g(x)>0$ or $g(x)<0$ when $x$ is extremely negative. To answer this, we can note that $$ g(x-1) = \ln\biggl( 1+\frac1{x-1} \biggr) - \frac1x = \ln\biggl( 1-\frac1x \biggr)^{-1} - \frac1x. $$ The function $\ln(1-u)^{-1}$ is convex (its second derivative is $(1-u)^{-2}$ which is positive) and therefore its values lie above its tangent lines; in particular, since its tangent line at $u=0$ is just $y=0$, we have $\ln(1-u)^{-1} > u$. Choosing $u=1/x$ confirms that $g(x-1)>0$ and thus $g(x)>0$. Indeed this proof works for all $x$ for which $g(x)$ is defined, not just $x$ extremely negative.

(Side note: $\ln(1-u)^{-1}$ is a particularly convenient function for estimations since its Maclaurin series coefficients are all positive.)

Answer 4

Notice that $\frac1{x+1}=\frac{1/x}{1+1/x}$ and thus

$$\begin{align} \lim_{x\to-\infty}\left(\ln\left(1+\frac1x\right)-\frac{1}{x+1}\right) &= \lim_{x\to0-}\left(\ln(1+x)-\frac{x}{1+x}\right) %&= \lim_{x\to0+}\left(\ln(1-x)+\frac{x}{1-x}\right) \end{align}$$ as $x$ is close to 0, we can expand either term as a power series:

$$ \ln(1+x) = x -\frac{x^2}2+\frac{x^3}3-\frac{x^4}4\pm\cdots $$ $$ -\frac x{1+x} = -x+x^2-x^3+x^4\mp\cdots $$ Adding these two series, the first non-vanishing term is $\displaystyle \frac{x^2}2$ which dominates. Hence the original term behaves like $0.5x^{-2}$ for large $x$.

Answer 5

Let $1+\frac1x=e^t.$ We have $$\lim_{x\to\pm\infty}\ln(1+\frac{1}{x})-\frac{1}{x+1}\\=\lim_{t\to 0}t-1+e^{-t}=0^+,$$ since $e^{-t}>1-t$, for $t\neq 0$.

Answer 6

One of my old professors used to claim "we are always closer to $0$ than to $\infty$"

$$x=\frac 1 y \qquad \implies \qquad \log(1+y)-\frac{y}{y+1}$$ Using the series for the logarithm and long division for the fraction gives $$\log(1+y)-\frac{y}{y+1}=\frac{y^2}{2}-\frac{2 y^3}{3}+O\left(y^4\right)$$