Thanks for catching that! I will make the corrections.

[karush]

$\tiny{242.ws8.d}$
$$\displaystyle
L_d=\lim_{x \to \infty}
\left[\frac{\arctan{(n)}}{\pi +\arctan{(n)}}\right]
=\frac{1}{3}$$

$\text{L' didn't work}$

 

[Greg]

That's not an indeterminate form, so L'Hopital's rule does not apply. The limit may be found by evaluating the numerator and denominator separately. Do you see how?

 

[Prove It]

$\tiny{242.ws8.d}$
$$\displaystyle
L_d=\lim_{x \to \infty}
\left[\frac{\arctan{(n)}}{\pi +\arctan{(n)}}\right]
=\frac{1}{3}$$

$\text{L' didn't work}$

First of all, this is not a series, as you aren't summing up terms. This is a sequence. Anyway

$\displaystyle \begin{align*} \frac{\arctan{(n)}}{\pi + \arctan{(n)}} &= \frac{1}{\frac{\pi}{\arctan{(n)}} + 1} \end{align*}$

What happens to the arctangent function as $\displaystyle \begin{align*} n \to \infty \end{align*}$?

 

[Greg]

This is a sequence.

That's odd - it looks like a limit to me - possibly part of a convergence test for a series that is not given. Why do you think it's a sequence?

 

[Prove It]

That's odd - it looks like a limit to me - possibly part of a convergence test for a series that is not given. Why do you think it's a sequence?

The convention is that when "n" is used as the variable, it's a sequence, while where "x" is used is a function.

 

[karush]

$\displaystyle \begin{align*} \frac{\arctan{(n)}}{\pi + \arctan{(n)}} &= \frac{1}{\frac{\pi}{\arctan{(n)}} + 1} \end{align*}$

What happens to the arctangent function as $\displaystyle \begin{align*} n \to \infty \end{align*}$?

$$\arctan(\infty)\implies\frac{\pi}{2}$$

$$\frac{1}{\frac{\pi}{\pi/2} + 1}
=\frac{1}{2+1}=\frac{1}{3}$$

 

[Greg]

$$\arctan(\infty)\implies\frac{\pi}{2}$$

Actually, I think you'd want to write

$$\lim_{n\to\infty}\arctan(n)=\frac{\pi}{2}$$

Note that $\tan(u)$ is one to one over the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and that $\tan(u)\to\infty$ as $u\to\frac{\pi}{2}$.

So,

$$\lim_{n\to\infty}\arctan(n)=\lim_{u\to\pi/2}\arctan(\tan u)=\frac{\pi}{2}$$

Also, you're writing

$$\lim_{x\to\infty}$$

in several places where the limit is with respect to $n$, so you should be writing

$$\lim_{n\to\infty}$$