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$f:[0,\infty)\to R, f(x)=\sqrt{x^2+x\ln{(e^x+1)}}$

I have this function and i need to find out the asymptotes to $+\infty$ (+infinity)

i calculate the horizontal ones, and they are $+\infty$

i can't calculate the limits at the obliques asymptotes

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  • $\begingroup$ $+\infty$ cannot be a vertical or horizontal asymptote. $\endgroup$ – Peter Foreman Jul 8 at 14:43
  • $\begingroup$ i know that they are not i mean that the limit is +∞ $\endgroup$ – xirunicole Jul 8 at 15:09
  • $\begingroup$ Where exactly did you get stuck? Can you show us your steps? $\endgroup$ – Michael Rybkin Jul 8 at 15:35
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If you want to find the oblique asymptote in the direction of positive infinity, then do this:

$$ a=\lim_{x\to+\infty}\frac{f(x)}{x}= \lim_{x\to+\infty}\frac{\sqrt{x^2+x\ln{(e^x+1)}}}{x}=\\ \lim_{x\to+\infty}\frac{\sqrt{x^2+x\ln{(e^x+1)}}}{\sqrt{x^2}}= \lim_{x\to+\infty}\sqrt{1+\frac{\ln{(e^x+1)}}{x}}=\\ \sqrt{1+1}=\sqrt{2}. $$

$$ b=\lim_{x\to+\infty}(f(x)-ax)= \lim_{x\to+\infty}\left(\sqrt{x^2+x\ln{(e^x+1)}}-\sqrt{2}x\right)=0 $$

The second limit is more difficult to calculate, but it equals $0$.

Plugging all this information into the equation $y=ax+b$, we get: $$y=\sqrt{2}x.$$

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  • $\begingroup$ can you explain why ln(e^x+1) / x is 1 ? please $\endgroup$ – xirunicole Jul 8 at 16:18
  • $\begingroup$ Again, you can use L'Hôpital's rule to find it. Do you know how to use it? If your problem is really just those limits, then post a separate question for each one asking how to evaluate them. And don't forget to explain where exactly you get stuck with them. $\endgroup$ – Michael Rybkin Jul 8 at 16:19
  • $\begingroup$ i know but for all the square i need to use it ? like it isn't take me so far $\endgroup$ – xirunicole Jul 8 at 16:24
  • $\begingroup$ I'm sorry, but I don't really understand what you're saying. I think you're asking whether L'Hôpital's rule should be used for the whole expressing including the square root? Correct? No. Only for $\frac{\ln(e^x+1)}{x}$. Because that's the limit you want to find at the moment. It just happens to be part of a bigger expression. $\endgroup$ – Michael Rybkin Jul 8 at 16:25
  • $\begingroup$ all this time i was thinking that i was needed to apply L'Hopital's rule for all √ - square but now i get it that i need to do it just for the fraction $\endgroup$ – xirunicole Jul 8 at 16:30
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Hint

When $x$ is large, $e^x$ is much larger then $e^x+1 \sim e^x$, then ....

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  • $\begingroup$ the problem is at $f(x) / x$ $\endgroup$ – xirunicole Jul 8 at 15:14
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Hint.

$$ \lim_{x \to +\infty} \frac {\sqrt {x^2 + x \ln (e^x + 1)}} x = \lim_{x \to +\infty} \sqrt {\frac {x^2 + x \ln (e^x + 1)} {x^2}} = \lim_{x \to +\infty} \sqrt {1 + \frac {\ln (e^x + 1)} x} $$

Can you go on from here?

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  • $\begingroup$ actually there i am, i dont know why ln(e^x+1) / x is 1 $\endgroup$ – xirunicole Jul 8 at 16:18

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