# Why is the horizontal asymptote of $\lim_{x \to \infty} \frac{x}{x^2+1}$ $y=0$?

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?

• that limits isn't equal to $1$. it's actually $0$. Jul 4, 2020 at 7:58
• 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). Jul 4, 2020 at 8:28

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 $$\infty$$.