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Let $f(x)=|\frac{\ln (|x|)}{1+\ln(|x|)}|\space\space\space\space\space\forall\space\space x\in\mathbb{R}\setminus\{-\frac 1e,0,\frac 1e\}$. I want to define this function.

So first I took the function:

$g(x)=\frac {\ln x}{1+\ln x}\space\space\space\forall\space\space x>0, x\neq\frac 1e$.

$g'(x)=\frac 1{x(1+ \ln x)^2}>0 \space\space\space\space\forall\space\space x>0 ,x\neq\frac 1e.$

so $g$ is strictly increasing and $\lim_{x\to0^+}g(x)=1=\lim_{x\to\infty}g(x).$

Now how do I compute the limits at $x=\frac 1e$ from both sides and why are they different? Also when I do limits of functions example when I had $\frac {\infty}{\infty}$ when $x\to0$ and $x\to\infty$, can I use l'hospital to compute them? And when I don't use L'hospital to compute such limits?

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Hint: make a substitution $t = \ln x$.

So calculate:

$$\lim_{x\to\frac 1{e}}\frac{\ln x}{1+\ln x}= \lim_{t\to -1}\frac{t}{1+t}$$

Now this limit clearly doesn't exist. Left limit is $+\infty $ and right $-\infty $.

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But $$\ln\left(\frac{1}{e}\right)=-1$$ and your denominator is Zero, you can not use the rules of L'Hospital. If $x$ tends to Zero we get

$\lim_{x\to 0}\frac{\frac{1}{x}}{\frac{1}{x}}=1$ .The same as $x$ tends to infinity.

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  • $\begingroup$ um I was talking about using L'hospital when $x\to0$ and $x\to\infty$. $\endgroup$ – C. Cristi Jul 7 '18 at 7:11
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First: Recall that $\ln(x)$ is monotonically increasing on its domain of $(0, \infty)$.

Next: Note that $\ln(1/e) = \ln(e^{-1}) = -1 \ln(e) = -1$.

When $x < 1/e$ by just a bit, $\ln(x) < -1$ by just a bit. And so approaching your limit of $1/e$ from the left, we find a numerator that is getting close to $-1$, and a denominator that is very close to $0$, but ever so slightly negative; so, the negatives cancel and yield a ratio that approaches $+\infty$.

When $x > 1/e$ by just a bit, $\ln(x) > -1$ by just a bit. And so approaching your limit of $1/e$ from the right, we find a numerator that is (again) getting close to $-1$, and a denominator that is very close to $0$, but ever so slightly positive; so, the $-1$ divided by something approaching $0$ but positive yields a ratio that approaches $-\infty$.

To see a graph of this function, as well as its vertical asymptote at $x = 1/e$, check here.

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We have that the numerator

  • $\ln x \to -1$

and the denominator

  • $1+\ln x \to 0$

but

  • $1+\ln x>0$ for $x\to \frac1e^+$

  • $1+\ln x<0$ for $x\to \frac1e^-$

therefore we can conclude that le limit doesn’t exist.

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