# Put limit in big O little O notation?

## So for the two functions, lnx and x-lnx. Compare which one grows faster and put it in the big o little o notation ( slower = o (faster) , top = O (bottom))

Jul 13, 2018

$\ln x = o \left(x - \ln x\right)$

#### Explanation:

Supposing we are here required to evaluate the behavior for $x \to \infty$, by definition, given two real functions $f \left(x\right)$ and $g \left(x\right)$:

${\lim}_{x \to \infty} \frac{f \left(x\right)}{g \left(x\right)} = 0 \iff f \left(x\right) = o \left(g \left(x\right)\right)$

and:

${\lim}_{x \to \infty} \text{sup} \left\mid \frac{f \left(x\right)}{g \left(x\right)} \right\mid < \infty \iff f \left(x\right) = O \left(g \left(x\right)\right)$

In our case let $f \left(x\right) = \ln x$ and $g \left(x\right) = x - \ln x$. Then:

${\lim}_{x \to \infty} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to \infty} \ln \frac{x}{x - \ln x}$

${\lim}_{x \to \infty} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to \infty} \frac{1}{\frac{x}{\ln} x - 1}$

Evaluate now the limit:

${\lim}_{x \to \infty} \frac{x}{\ln} x$

it is in the indeterminate form $\frac{\infty}{\infty}$ so we can use l'Hospital's rule:

${\lim}_{x \to \infty} \frac{x}{\ln} x = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \left(x\right)}{\frac{d}{\mathrm{dx}} \left(\ln x\right)}$

${\lim}_{x \to \infty} \frac{x}{\ln} x = {\lim}_{x \to \infty} \frac{1}{\frac{1}{x}} = {\lim}_{x \to \infty} x = + \infty$

Then:

${\lim}_{x \to \infty} \frac{1}{\frac{x}{\ln} x - 1} = 0$

which means:

$\ln x = o \left(x - \ln x\right)$