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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!

enter image description here

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  • $\begingroup$ I would say $\ln(x) \lt \dfrac{\sqrt{x}}{2}$ for large $x$ $\endgroup$ – Henry Oct 4 '16 at 16:11
  • $\begingroup$ @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... $\endgroup$ – imranfat Oct 4 '16 at 16:18
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Hint: $$ x-\sqrt{x}\,\ln x=x\,\Bigl(1-\frac{\ln x}{\sqrt x}\Bigr). $$

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  • $\begingroup$ 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. $\endgroup$ – Bib-lost Oct 4 '16 at 16:48
  • $\begingroup$ Hi. Thanks for the suggestion! $\endgroup$ – FoolishNumber Oct 4 '16 at 17:14
  • $\begingroup$ I think I know how to do it. However, I just want to double check(Sorry). So from my understanding, I should be able to solve for the limit of lnx/root(x) going to infinity, which equals 0. Then I would plug that in for your suggested equation after plugging in x = infinity. That would lead to me having infinity as the answer? Again, Thank you! This is what I did if you want to see. [puu.sh/rxC0I/d186203698.jpg] $\endgroup$ – FoolishNumber Oct 4 '16 at 17:24
  • $\begingroup$ Yes, taht's right. $\endgroup$ – Julián Aguirre Oct 5 '16 at 16:52

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