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# limits questions

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How can I do these Oct 24, 2014

#8
+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

Oct 24, 2014

#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}$$I know pi/2 is right but I am really confused dismissing other answers. This answer is DEFINITELY NOT completely correct. Oct 24, 2014 #2 0 My argument for #18: 1) Since lim(x→∞) (x² + 1)/(x + 1) = ∞, 2) the problem has the same limit as lim(x→∞) invSec(x) 3) which has the same limit as lim(x→∞) invCos(1/x) 4) and since lim(x→∞) (1/x) = 0, 5) it becomes lim(x→0) Cos(x) = 1. Any questions on any of the step? I haven't looked at the other two yet. Oct 24, 2014 #3 +5 20) I drew a triangle to help me with this.$$\\\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

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Oct 24, 2014
#4
0

Ignore my answer for #18; I messed up!

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.

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

Oct 24, 2014
#7
+5 .

Oct 24, 2014
#8
+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
#9
0

sorry but I cant understand 18

Oct 24, 2014