Question

Let F(x)=ln(1+x^2)
A) limit as x approaches infinity
B)F1(x) as it approaches infinity
C) explain the behavior as function x gets large

Answer 1

what is \(\ln(1+\infty^2)\)?

Answer 2

you mean the derivative?

Answer 3

no we are just doing limits, not derivatives. Part asks you what happens to \(\ln(1+x^2)\) when x gets very large. So what happens when you "plug in" infinity?

Answer 4

Part A asks*

Answer 5

the number becomes particularly small ?

Answer 6

for example ln (1+1000^2)=13.8155

Answer 7

make it 10^50

Answer 8

its actually getting larger, so it will tend to infinity. VERY SLOWLY......

Answer 9

115.129

Answer 10

no try 10 ^1000000000000000000000

Answer 11

now try*

Answer 12

my calculator wont allow me but i guess its bc it is very large

Answer 13

yes, it keeps getting larger

Answer 14

what is F1(x)? Do you mean \(f^{-1}(x)\)?

Answer 15

no first derivative as it approaches infinity

Answer 16

ahh good, otherwise we were in trouble.

Answer 17

do you know the first derivative?

Answer 18

so for part A i would put infinity ?

Answer 19

yes, or undefined

Answer 20

2x/x^2+1

Answer 21

that is the derivative

Answer 22

sorry, phone. ok do you know how to take the limit?

Answer 23

it confuses me a little ?

Answer 24

my teacher alwasy taught us to just plug it in

Answer 25

well you have \(\frac{2x}{x^2+1}\) if you divide every term by x you get \(\frac{2}{x+\frac{1}{x}}\) if you take the limit as x goes to infinity you get \(\frac{2}{\infty+\frac{1}{\infty}}=\frac{2}{\infty+0}=\frac{2}{\infty}=?\)

Answer 26

a very small number

Answer 27

approaches zero ?