Wolfram Alpha evaluates this limit
$$\lim_{x\to \infty } \, \frac{x}{\ln (x)-\ln \left(\frac{1}{x}\right)}$$
to be infinity.
But I suspect it could be a real number. What is the correct answer?
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Wolfram alpha is right because the denominator is $\ln(x)-\ln(\frac1x)=2\ln x$, which grows slower than $x$.
Hagen von Eitzen
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1$y = 2x$ grows slower than $y = 3x$ but $\lim_{x \to \infty} \frac{3x}{2x}$ is not $\infty.$ – DDS Jun 30 '19 at 03:21
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$$\lim_{x \to \infty} \frac{x}{\ln x - \ln x^{-1}} = \lim_{x \to \infty} \frac{x}{\ln x + \ln x} = \lim_{x \to \infty} \frac{x}{2 \ln x}$$
which gives the indeterminate form $\frac{\infty}{\infty}$, and so applying L'Hospital's Rule we get
$$ \lim_{x \to \infty} \frac{1}{2\left( \frac{1}{x} \right)} = \lim_{x \to \infty} \frac{x}{2} = \infty.$$
DDS
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$ln(a)-ln(b)=ln(a/b)$ so $ln(x)-ln(1/x)=ln(x/(1/x))$ which is $ln(x^2)$ which is equal to $2*ln(x)$ and then the original equation becomes $x/{(2*ln(x))}$ which using l'hospital's rule becomes $1/(2*1/x)$ which simplifies to $x/2$ and the limit as that aproaches infinity is clearly infinity.
