# What's the limit of xArctan(x/x^2+1) as x approaches positive infinity ?

Jul 25, 2017

${\lim}_{x \to \infty} x \arctan \left(\frac{x}{{x}^{2} + 1}\right) = 1$

#### Explanation:

The limit:

${\lim}_{x \to \infty} x \arctan \left(\frac{x}{{x}^{2} + 1}\right)$

is in the indeterminate form $+ \infty \times 0$. We can reconduce it to the form $\frac{0}{0}$ and apply l'Hospital's rule:

${\lim}_{x \to \infty} x \arctan \left(\frac{x}{{x}^{2} + 1}\right) = {\lim}_{x \to \infty} \arctan \frac{\frac{x}{{x}^{2} + 1}}{\frac{1}{x}}$

${\lim}_{x \to \infty} x \arctan \left(\frac{x}{{x}^{2} + 1}\right) = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \arctan \left(\frac{x}{{x}^{2} + 1}\right)}{\frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)}$

Now evaluate:

$\frac{d}{\mathrm{dx}} \arctan \left(\frac{x}{{x}^{2} + 1}\right) = \frac{1}{1 + {\left(\frac{x}{{x}^{2} + 1}\right)}^{2}} \frac{d}{\mathrm{dx}} \left(\frac{x}{{x}^{2} + 1}\right)$

$\frac{d}{\mathrm{dx}} \arctan \left(\frac{x}{{x}^{2} + 1}\right) = {\left({x}^{2} + 1\right)}^{2} / {\left(x + \left({x}^{2} + 1\right)\right)}^{2} \frac{\left({x}^{2} + 1\right) - 2 {x}^{2}}{{x}^{2} + 1} ^ 2$

$\frac{d}{\mathrm{dx}} \arctan \left(\frac{x}{{x}^{2} + 1}\right) = - \frac{{x}^{2} - 1}{x + \left({x}^{2} + 1\right)} ^ 2$

$\frac{d}{\mathrm{dx}} \arctan \left(\frac{x}{{x}^{2} + 1}\right) = - \frac{{x}^{2} - 1}{{x}^{4} + 2 {x}^{2} + x + 1}$

Then:

${\lim}_{x \to \infty} x \arctan \left(\frac{x}{{x}^{2} + 1}\right) = {\lim}_{x \to \infty} \frac{- \frac{{x}^{2} - 1}{{x}^{4} + 2 {x}^{2} + x + 1}}{- \frac{1}{x} ^ 2}$

${\lim}_{x \to \infty} x \arctan \left(\frac{x}{{x}^{2} + 1}\right) = {\lim}_{x \to \infty} {x}^{2} \left(\frac{{x}^{2} - 1}{{x}^{4} + 2 {x}^{2} + x + 1}\right)$

${\lim}_{x \to \infty} x \arctan \left(\frac{x}{{x}^{2} + 1}\right) = {\lim}_{x \to \infty} \left(\frac{{x}^{4} - {x}^{2}}{{x}^{4} + 2 {x}^{2} + x + 1}\right)$

${\lim}_{x \to \infty} x \arctan \left(\frac{x}{{x}^{2} + 1}\right) = 1$

graph{x arctan(x/(x^2+1)) [-10, 10, -5, 5]}