Determine $\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}(\sqrt{n+1})-\mathrm{arctan}(\sqrt{n})$

Jazmine Bruce
2022-05-18
Answered

Determine $\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}(\sqrt{n+1})-\mathrm{arctan}(\sqrt{n})$

You can still ask an expert for help

Julien Carrillo

Answered 2022-05-19
Author has **13** answers

Note,

$\mathrm{arctan}\sqrt{n+1}-\mathrm{arctan}\sqrt{n}=\mathrm{arctan}\frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n+1}\cdot \sqrt{n}}$

$=\mathrm{arctan}\frac{1}{(1+\sqrt{{n}^{2}+n})(\sqrt{n+1}+\sqrt{n})}$

Thus,

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}(\sqrt{n+1})-\mathrm{arctan}(\sqrt{n})$

$=\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}\frac{1}{(1+\sqrt{{n}^{2}+n})(\sqrt{n+1}+\sqrt{n})}=\mathrm{arctan}(0)=0$

$\mathrm{arctan}\sqrt{n+1}-\mathrm{arctan}\sqrt{n}=\mathrm{arctan}\frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n+1}\cdot \sqrt{n}}$

$=\mathrm{arctan}\frac{1}{(1+\sqrt{{n}^{2}+n})(\sqrt{n+1}+\sqrt{n})}$

Thus,

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}(\sqrt{n+1})-\mathrm{arctan}(\sqrt{n})$

$=\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}\frac{1}{(1+\sqrt{{n}^{2}+n})(\sqrt{n+1}+\sqrt{n})}=\mathrm{arctan}(0)=0$

Nicholas Cruz

Answered 2022-05-20
Author has **3** answers

The arctan function is the inverse function of

$\mathrm{tan}:(-\frac{\pi}{2},\frac{\pi}{2})\to \mathbb{R}$

as this function is monotonically increasing, we have

$\underset{x\to \frac{\pi}{2}}{lim}\mathrm{tan}x=+\mathrm{\infty}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\underset{x\to +\mathrm{\infty}}{lim}\mathrm{arctan}x=\frac{\pi}{2}$

therefore

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}(\sqrt{n+1})=\frac{\pi}{2}$

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}(\sqrt{n})=\frac{\pi}{2}$

so the difference is 0.

$\mathrm{tan}:(-\frac{\pi}{2},\frac{\pi}{2})\to \mathbb{R}$

as this function is monotonically increasing, we have

$\underset{x\to \frac{\pi}{2}}{lim}\mathrm{tan}x=+\mathrm{\infty}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\underset{x\to +\mathrm{\infty}}{lim}\mathrm{arctan}x=\frac{\pi}{2}$

therefore

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}(\sqrt{n+1})=\frac{\pi}{2}$

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{arctan}(\sqrt{n})=\frac{\pi}{2}$

so the difference is 0.

asked 2020-12-16

The figure shows the surface created when the cylinder ${y}^{2}+{Z}^{2}=1$ intersects the cylinder ${x}^{2}+{Z}^{2}=1$. Find the area of this surface.

The figure is something like:

asked 2022-04-14

How do you find the average rate of change of $f\left(x\right)=\frac{2}{{x}^{2}}$ over the interval [1, 5]?

asked 2022-02-12

How does tangent slope relate to the slope of a line?

asked 2022-04-13

How do you find the average rate of change of f(t)=2t+7 from [1,2]?

asked 2021-12-12

What is the derivative of x?

asked 2022-04-14

Find the limit of the complex function.

$\underset{x\to a}{lim}{(2-\frac{x}{a})}^{\left(\mathrm{tan}\frac{\pi x}{2a}\right)}$

I have simplified this limit to this extent :

$e}^{\underset{x\to a}{lim}\left((1-\frac{x}{a})\left(\mathrm{tan}\frac{\pi x}{2a}\right)\right)$

I don't know how to simplify the limit after that.

I have simplified this limit to this extent :

I don't know how to simplify the limit after that.

asked 2022-04-30

I have an integral here that I'm trying to figure out.

$\int 7{\mathrm{sin}}^{2}x{\mathrm{cos}}^{4}xdx$