Comparing the asymptotic growth rate of functions $y= \log(x)+ x$ and $y = x^{0.99}\log(x)$

Question

Using the limits criteria i.e. $\lim_{x \to \infty} f(x)/g(x)$, I found that $x + \log(x)$ grows faster than $\log(x)*x^{0.99}$. However the graph is quite contradictory to what I evaluated.

Graph of $x^{0.99}\log(x)$ and $\log(x)+ x$

Why am I getting this discrepancy?

Answer 1

Your graph is nice, however $10^{11}$ is too small number to see that $x+\log x$ grows faster. It starts to do that from approx. $10^{281}$, see here.

Answer 2

Yes you are right indeed we have that

$$\frac{x+\log x}{x^{0.99}\log x}=\frac{x^{0.01}}{\log x}+\frac{1}{x^{0.99}}\to \infty$$

the discrepancy is due to the very slow rate of divergence of the ratio.

Answer 3

As your graph suggests, and your calculations show we have that although $f(x) = x+\log(x)$ grows asymptotically faster than $g(x)=x^{0.99}\log(x)$, the latter is actually much larger for values at least up to $10^{12}$.

So let's try to solve $f(x) = g(x)$, i.e. find where the two graphs meet. However, for the given functions this is algebraically too difficult, so instead let's look at the functions $$\hat f(x) = x, \qquad \hat g(x) = x^{1-\epsilon}\log(x)$$ (the $\log(x)$ term is more or less irrelevant for the asymptotic behaviour anyway)

Setting $\hat f(x) = \hat g(x)$, substituting $x=e^y$ and simplifying yields:

$$ 1 = e^{-\epsilon y}y$$

which we can solve using the lambert-W function (check out the wiki page) giving $$y = e^{-W_{-1}(-\epsilon)}$$

Now, if we look at the graph of the lambert-W function you will see that it has two branches. For us the green one is the important one. As you can see as we approach zero from the left there is a singularity, which means that for small $\epsilon$, the exponent $-W_{-1}(-\epsilon)$ is a large positive number. But the point at which the curves intersect is given by $$x = e^{e^{-W_{-1}(-\epsilon)}}$$ and hence must be astronomically huge.

Answer 4

A trick to finding the asymptopic behaviour of functions with small differences is to take the log of both functions and plot them (or if need be, multiple logs). Since the log function is always increasing and injective, it does not change the relative value of each function. For your example, taking the log finds that the first function overtakes the other around the thousands, rather than on the order of 10^281.