I figured, the limit of below is:

$$\lim_{x \to \infty} \frac{x}{x^2+1} = \frac{\infty}{\infty^2 + 1} \approx \frac{\infty}{\infty} = 1$$.

Should the horizontal asymptote be $y=1$ then?

  • 1
    $\begingroup$ that limits isn't equal to $1$. it's actually $0$. $\endgroup$
    – alphaomega
    Jul 4, 2020 at 7:58
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    $\begingroup$ You can see it like: "the higher exponent dominates". Your $x^2$ in the denominator grows a lot faster than your $x$ in the numerator, so when going to infinity, you can ignore the lower exponent $x$ because $x^2$ will be much much higher, leading to a result equal to 0 (This is not the best quality explanation, check the mathematical operations people have written in the answers, but may work to imagine how this kind of limits work). $\endgroup$ Jul 4, 2020 at 8:28

1 Answer 1


You have $$\lim_{x\rightarrow\infty}\frac{x}{x^2+1}=\lim_{x\rightarrow\infty}\frac{1}{x+\frac{1}{x}}=0$$ which You see after multiplying by $\frac{1}{x}$ in enumerator and denominator. It easily leads to false results if You carlessly use the symbol $\infty$.


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