# Limit of $x - \sqrt{x} \ln x$ as $x$ approaches infinity

The problem: $$\lim_{x\to\infty}x-\sqrt{x}\ln x$$

I can't seem to figure out how to find the limit of this problem.

Can someone guide or help me solve the problem?

What I tried so far

I tried multiplying top and bottom by $x + \sqrt{x}\ln x$ to cancel out the roots. However, it ends up as (in the picture) which when plugging in infinity, doesn't amount to anything ( maybe I'm wrong). I'm confused on what other method to do next as typically to solve these type of problems, you would multiply top and bottom by its opposite, which I tried. Thank you for any assistance! • I would say $\ln(x) \lt \dfrac{\sqrt{x}}{2}$ for large $x$ – Henry Oct 4 '16 at 16:11
• @HiDanny To follow your approach (I did the work), you would have to do L'Hospital's Rule 4 times until finally the denominator get "exhausted" to see the the numerator wins, →infinity. But there are more easier and elegant approaches, like the suggestions here... – imranfat Oct 4 '16 at 16:18

Hint: $$x-\sqrt{x}\,\ln x=x\,\Bigl(1-\frac{\ln x}{\sqrt x}\Bigr).$$
• If $\lim_{x \to +\infty} \frac{\ln x}{\sqrt{x}}$ is anything but $1$, we immediately get the limit from the original question by applying the product rule for limits. Remains to calculate $\lim_{x \to +\infty} \frac{\ln x}{\sqrt{x}}$, which can, for example, be done by using l'Hôpital. – Bib-lost Oct 4 '16 at 16:48