Calculate the limit using de L'Hopital's rule

Question

Calculate the following limit:

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

I started like this:

$\lim_{x \to +\infty}(\sqrt{x}-\log x)=[\infty-\infty]=\lim_{x \to +\infty}\frac{(x-(\log x)^2)}{(\sqrt{x}+\log x)}=$

but that's not a good way...

I would be gratefull for any tips.

Do you really need L'Hospital's rule for this limit? It's a high-school exercise. — Bernard, Dec 08 '17 at 14:19
You just need to take $\sqrt{x} $ as a factor and use $(\log x) /\sqrt{x} \to 0$ as $x\to\infty$. — Paramanand Singh, Dec 08 '17 at 14:41

StackTD · Accepted Answer · 2017-12-08T14:05:13.587

4

Hint

If you have to use l'Hôpital; this limit is easier to find (*): $$\lim_{x \to +\infty} \frac{\sqrt{x}}{\log x} = \lim_{x \to +\infty} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{x}} =\lim_{x \to +\infty}\frac{\sqrt{x}}{2} = +\infty$$ Can you see how this would help for your limit as well?

If not (hoover over), rewrite:

$$\sqrt{x}-\log x = \left( \frac{\sqrt{x}}{\log x} - 1 \right) \log x$$

(*) With a similar calculation, it's easy to show and worth remembering that for $n>0$: $$\lim_{x \to +\infty} \frac{x^n}{\log x} = +\infty$$

edited Dec 08 '17 at 14:05

answered Dec 08 '17 at 13:58

StackTD

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Thanks. So I have: $\sqrt(x)-log(x)=(\frac{\sqrt(x)}{logx}-1)log(x)=\frac{\frac{\sqrt(x)}{logx}-1}{1/log(x)}$ and it is $+\infty/ 0$, I still don't know what is next – SigmaMat Dec 08 '17 at 14:14
No need to go further than $\left( \frac{\sqrt{x}}{\log x} - 1 \right) \log x$ since (obviously) $\log x \to +\infty$ and by our earlier l'Hôpital, also $\frac{\sqrt{x}}{\log x} \to +\infty$; you see? – StackTD Dec 08 '17 at 14:20
Ok, we get two factors both going to $+\infty$ so the product of them goes to $+\infty$. Right? Thanks a lot! – SigmaMat Dec 08 '17 at 14:28
Correct! And you're welcome. – StackTD Dec 08 '17 at 14:28

score 2 · Answer 2 · edited Dec 08 '17 at 14:08

$\lim_{x \to +\infty}(\sqrt{x}-\log(x))=[\infty-\infty]=\lim_{x \to +\infty}\frac{(x-\log(x)^2)}{(\sqrt{x}+\log(x))}=\lim_{x \to +\infty}\frac{\sqrt x-\frac{log(x)^2}{\sqrt x}}{\frac{\log x}{\sqrt x}+1}$

$-\frac{\log(x)^2}{\sqrt x}$ and $\frac{\log(x)}{\sqrt x}$ both tend to zero as $x$ tends to $\infty$.

Then your limit is $\lim_{x \to \infty}\sqrt x=+\infty$.

score 0 · Answer 3 · answered Dec 08 '17 at 16:59

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

$$\lim_{x \to +\infty} \left(\dfrac{\sqrt{x}-\log x}{1} \cdot \dfrac{\sqrt{x}+\log x}{\sqrt{x}+\log x} \right)$$

$$\lim_{x \to +\infty} \left( \dfrac{x-(\log x)^2}{\sqrt{x}+\log x} \right)$$

$$\lim_{x \to +\infty} \left( \dfrac{1-\frac{2\log x}{x}}{\frac{1}{2\sqrt x}+\frac 1x} \right) \to \dfrac 10\to \infty$$

Calculate the limit using de L'Hopital's rule

3 Answers3