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How do I show that $\lim_{x \rightarrow \infty } \frac {\log(x^{2}+1)}{x}=0$?I was able to do that using L'Hôpital's rule. But is there any other way?

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you can use fact that denominator is going increase more rapidly then numerator –  dato datuashvili Jan 12 at 12:39
    
yes, but how do I show that $x$ increases faster than $log(x^{2}+1)$. –  D.N Jan 12 at 12:40
    
@cryogenic My hint, though correct, was misleading. I got confused with something. I deleted my answer. –  Git Gud Jan 12 at 12:50
    
@cryogenic I fixed my answer. –  Git Gud Jan 12 at 12:56
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5 Answers

up vote 2 down vote accepted

$$\lim_{x\to\infty}\frac{\ln(x^2+1)}x\sim\lim_{x\to\infty}\frac{\ln(x^2)}x=\lim_{x\to\infty}\frac{2\cdot\ln x}x=2\cdot\lim_{t\to\infty}\frac t{e^t}=2\cdot0=0.$$

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between the first two limit there should be a $=$; the limits are the same :-) –  Ant Jan 12 at 22:40
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See as the limit is of (infinity / infinity ) form we can use L'hospital rule on this limit..... we have to apply the derivative on the numerator and denominator repeatedly untill we have a fraction is of form other than (infinity / infinity )...... Then we apply the process for two time and we have a fraction as: lim(x--->infinity) (2) / (2x + 1) . and now we put x--> infinity in the limit.....and we have the limiting value as ZERO or 0.

Hope this discussion will help you.....best of luck.

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You may write directly, as $\lim_{x \rightarrow \infty} \frac{\log (x)}{x^a}=0$, $a>0$ $$\frac{\log (1+ x^2)}{x} \cong \frac{\log (x^2)}{x} =\frac{2\log (x)}{x} \overset{ x \rightarrow \infty}{\longrightarrow} 0$$

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You can always brute force it by $\log(x^2+1)\leq \log 2 + 2\log x$ (for $x\geq 1$), replace $y=\log x$ and use a Taylor expansion on $\exp(y)$.

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This is my first semester of analysis(/calculus?) and I don't know how to use Taylor expansion yet. Thank you for your answer though. –  D.N Jan 12 at 13:03
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Hint: Set $x^2=e^t-1$ to get $\lim \limits_{x\to +\infty}\left(\dfrac{\log(x^2+1)}{x}\right)=\lim \limits_{t\to +\infty}\left(\dfrac{t}{\sqrt{e^t-1}}\right)$.

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Thank you, this is clearer to me now :) –  D.N Jan 12 at 12:58
    
@cryogenic You're welcome. Let me suggest you don't accept my answer so fast. That way you may discourage people from posting alternate solutions which you might even like better. –  Git Gud Jan 12 at 12:59
    
I will bear that in mind. –  D.N Jan 12 at 13:01
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