# Evaluate limit of $x\left|{\log{\left(x+1\right)}}\right|$ for $x\rightarrow(-1)^+$

I want to evaluate the following function:

$$x\left|{\log{\left(x+1\right)}}\right|$$

for $x\rightarrow(-1)^+$.

This is what I'm doing. Since $\left|{\log{\left(x+1\right)}}\right|$ is a negative quantity that approaches $-\infty$ for $x\rightarrow(-1)^+$ we have that:

$$x\left|{\log{\left(x+1\right)}}\right|=-x\log{\left(x+1\right)}$$

Now we have that $x+1\rightarrow0$ for $x\rightarrow(-1)^+$ therefore $\log{\left({x+1}\right)}\sim x+1$

This should mean that

$$f(x)\sim -x\left({x+1}\right)\sim -(-1)(x+1)=x+1\rightarrow0^-$$

However my texbook reports that the limit is actually $-\infty$. Any hints on what I did wrong?

The following does not hold : $\log x \sim x$ as $x \to 0$.

Now, the limit should be :

$$\lim_{x \to -1^+} x|\log(x+1)| = \lim_{x \to -1^+} -|\log(x+1)| = - \infty$$

since $\log(x+1)$ tends to $-\infty$ for $(x+1) \to 0$.

\begin{align} \lim_{x\to(-1)^+}x|\ln (1+x)|&=\lim_{\epsilon\to 0^+}(\epsilon -1)|\ln(\epsilon-1+1)|\\ &=\lim_{\epsilon\to 0^+}(\epsilon -1)|\ln\epsilon|\\ &=-\infty \end{align}

It is not true that $\log t \sim t$ as $t \to 0$. To find $\lim_{x\to (-1)^{+}} -x\log (1+x)$ note that $\log (1+x) \to -\infty$ and $-x \to 1$. Just multiply the limits.

• This should be a comment, not an answer. – uniquesolution Aug 22 '18 at 8:09
• @uniquesolution The OP question was "Any hints on what I did wrong ?". So that's an eligible answer. – Rebellos Aug 22 '18 at 8:10
• @Rebellos You are entitled to your opinion. – uniquesolution Aug 22 '18 at 8:11
• Thanks, I didn't see that. This hint helped me – Cesare Aug 22 '18 at 8:11

An option:

$y:= \log (1/(x+1)), y >0$ for

$0> x >-1,$ then

$\lim_{(x+1) \rightarrow 0^+} y= \infty.$

With $e^y:=1/(x+1)$,

$|\log (x+1)|= |\log (1/(x+1))| = |\log e^y|= y.$

$\lim_{(x+1) \rightarrow 0^+} x|\log(1+x)|=$

$\lim_{y \rightarrow \infty}(e^{-y}-1)y= -\infty.$