+0  
 
0
445
9
avatar+1832 

How can I do these

 

xvxvxv  Oct 24, 2014

Best Answer 

 #8
avatar+17736 
+5

My argument for #22:

As x→0+, ln(x) → -∞.

When does tan(x) → -∞?

As x → -π/2 (from the plus side), tan(x) → -∞.

So, as tan(x) → -∞ (from the plus side), x → -π/2 

So, as x → -∞, invTan(x) →  -π/2 

So, the limit is  -π/2 

geno3141  Oct 24, 2014
 #1
avatar+92624 
+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
avatar+17736 
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
avatar+92624 
+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
 #4
avatar+17736 
0

Ignore my answer for #18; I messed up!

geno3141  Oct 24, 2014
 #5
avatar+17736 
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
avatar+17736 
+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
 #7
avatar+26720 
+5

limits

 

.

Alan  Oct 24, 2014
 #8
avatar+17736 
+5
Best Answer

My argument for #22:

As x→0+, ln(x) → -∞.

When does tan(x) → -∞?

As x → -π/2 (from the plus side), tan(x) → -∞.

So, as tan(x) → -∞ (from the plus side), x → -π/2 

So, as x → -∞, invTan(x) →  -π/2 

So, the limit is  -π/2 

geno3141  Oct 24, 2014
 #9
avatar+1832 
0

sorry but I cant understand 18 

xvxvxv  Oct 24, 2014

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