#1**+5 **

Hi 315,

Alan can you check this please.

I am not very good at limits but I will give it a shot.

$$\\0 so\\

\infty>sec^{-1}x \qquad \;\;and\;\; sec^{-1}x>\frac{1}{\pi}\\\\

\frac{1}{\pi}

$$\lim\limits_{x\rightarrow \infty}\;\;sec^{-1}\;\;\frac{x^2+1}{x+1}\\\\

=\lim\limits_{x\rightarrow \infty}\;\;sec^{-1}\;\;\frac{x+\frac{1}{x}}{1+\frac{1}{x}}\\\\

=sec^{-1}\;\;\frac{\infty}{1}\\\\

=\frac{\pi}{2}$$

I know pi/2 is right but I am really confused dismissing other answers.

This answer is DEFINITELY NOT completely correct.

Melody Oct 24, 2014

#2**0 **

My argument for #18:

1) Since lim(x→∞) (x² + 1)/(x + 1) = ∞,

2) the problem has the same limit as lim(x→∞) invSec(x)

3) which has the same limit as lim(x→∞) invCos(1/x)

4) and since lim(x→∞) (1/x) = 0,

5) it becomes lim(x→0) Cos(x) = 1.

Any questions on any of the step?

I haven't looked at the other two yet.

geno3141 Oct 24, 2014

#3**+5 **

20) I drew a triangle to help me with this.

$$\\\lim\limits_{x\rightarrow\infty}\;\;Sin(tan^{-1}x)\\\\

\lim\limits_{x\rightarrow\infty}\;\;\frac{x}{\sqrt{1+x^2}}\\\\

\lim\limits_{x\rightarrow\infty}\;\;\frac{x}{\sqrt{x^2(\frac{1}{x^2}+1})}\\\\

\lim\limits_{x\rightarrow\infty}\;\;\frac{x}{x\sqrt{\frac{1}{x^2}+1}}\\\\

\lim\limits_{x\rightarrow\infty}\;\;\frac{1}{\sqrt{\frac{1}{x^2}+1}}\\\\

=\frac{1}{\sqrt{0+1}}\\\\

=1$$

Melody Oct 24, 2014

#5**0 **

I will go back to my original analysis, except for the last step; it has a limit of π/2, not 1.

geno3141 Oct 24, 2014

#6**+5 **

My argument for #20:

As x → π/2, tan(x) → ∞.

So, as x → ∞, invTan(x) → π/2.

As x → π/2, sin(x) → 1

geno3141 Oct 24, 2014