Saturday, August 11, 2018

calculus - Limit of $x - ln(x)$ as $x$ approaches $+infty$



To evaluate the limit of an even larger expression



$$
\lim_{x \to +\infty} \frac{\ln(\ln x)}{\ln(x - \ln x)}
$$




I need to evaluate part of the denominator to determine whether I could apply L'Hôpital's Rule



$$
\lim_{x \to +\infty} x - \ln(x)
$$



The problem is that I can't seem to manipulate the expression to the indeterminate forms $0/0$ or $\infty/\infty$. I was thinking of multiplying by $x/x$



$$
\lim_{x \to +\infty} \frac{x^2 - x\ln(x)}{x}

$$



But then I get another indeterminate form $\infty - \infty$ in the numerator. I was also thinking that $x$ grows much faster than $\ln x$, so the limit obviously tends to $+\infty$ but I don't think that would fly with most people :)


Answer



$x-\ln{x}=\ln{e^x}-\ln{x}=\ln{\frac{e^x}{x}}\,,$ and $\frac{e^x}{x}$ is in the form you want.


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