# Obliques asymptotes of a function

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

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

• $+\infty$ cannot be a vertical or horizontal asymptote. – Peter Foreman Jul 8 '19 at 14:43
• i know that they are not i mean that the limit is +∞ – xirunicole Jul 8 '19 at 15:09
• Where exactly did you get stuck? Can you show us your steps? – Michael Rybkin Jul 8 '19 at 15:35

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: $$y=\sqrt{2}x.$$

• can you explain why ln(e^x+1) / x is 1 ? please – xirunicole Jul 8 '19 at 16:18
• 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. – Michael Rybkin Jul 8 '19 at 16:19
• i know but for all the square i need to use it ? like it isn't take me so far – xirunicole Jul 8 '19 at 16:24
• 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. – Michael Rybkin Jul 8 '19 at 16:25
• 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 – xirunicole Jul 8 '19 at 16:30

Hint

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

• the problem is at $f(x) / x$ – xirunicole Jul 8 '19 at 15:14

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?

• actually there i am, i dont know why ln(e^x+1) / x is 1 – xirunicole Jul 8 '19 at 16:18