Find lim as x to infinity (x+ arctan x)/(x³+x²+x)

Find $\displaystyle\lim_{x\to\infty}\frac{x+\arctan x}{x^{3}+x^{2}+x}$.

Step 1: Identify the dominant terms

As $x\to\infty$ we have $\arctan x\to\dfrac{\pi}{2}$, a bounded quantity, so the numerator behaves like $x$. The denominator is dominated by $x^{3}$. Both tend to $+\infty$, so the form is $\dfrac{\infty}{\infty}$.

Step 2: Divide by the dominant power

Divide numerator and denominator by $x^{3}$: \[\frac{x+\arctan x}{x^{3}+x^{2}+x} = \frac{\dfrac{1}{x^{2}}+\dfrac{\arctan x}{x^{3}}}{1+\dfrac{1}{x}+\dfrac{1}{x^{2}}}.\]

Step 3: Evaluate the limit

Each term of the numerator tends to zero; the middle one because $\arctan x$ never exceeds $\pi/2$: \[\frac{1}{x^{2}}\to0, \qquad 0\lt \frac{\arctan x}{x^{3}}\le\frac{\pi/2}{x^{3}}\to0 \quad (x\gt 0),\]

while in the denominator $\dfrac1x\to0$ and $\dfrac{1}{x^{2}}\to0$. Hence

\[ \lim_{x\to\infty}\frac{\dfrac{1}{x^{2}}+\dfrac{\arctan x}{x^{3}}}{1+\dfrac{1}{x}+\dfrac{1}{x^{2}}} = \frac{0+0}{1+0+0} = 0. \]

Step 4: Verification

\[x = 10:\ \frac{11.4712}{1110} = 0.0103343; \qquad x = 100:\ \frac{101.5608}{1010100} = 1.00545\cdot10^{-4} \;\;\checkmark\]

\[x = 10^{3}:\ 1.00057\cdot10^{-6}; \qquad x = 10^{6}:\ 1.0000\cdot10^{-12} \;\;\checkmark\]

\[\text{The values follow } \frac{1}{x^{2}} \text{ closely, as the leading behaviour } \frac{x}{x^{3}} \text{ predicts} \;\;\checkmark\]