Question

For which pair of functions f(x) and g(x) below will the lim x-> infinity f(x)g(x) NOT EQUAL 0
f(x) = 10x + e-x; g(x) =

f(x) = x2; g(x) = e-4x

f(x) =(Lnx)3; g(x) =

f(x) = ; g(x) = e-x

Answer 1

I think that a or c could work so how do you determine which (if thats even correct)

Answer 2

I graphed them all

Answer 3

Do you know the L'Hospital's rule?

Answer 4

and a and c are the only ones that do not approach 0

Answer 5

not really

Answer 6

For a, we would set
\(\color{black}{\displaystyle \lim_{x\to\infty}(10x)(e^{-x})=\lim_{x\to\infty}\frac{10x}{e^x} }\)

Answer 7

You don't even need the L'Hospital's rule, you can just note that from roughly \(x=6\) and on, \(2^x>10x\), so a larger sequence would be \(2^x/e^x=(2/e)^x\).

Answer 8

Well, we know that \(\color{black}{\displaystyle \lim_{x\to\infty}\frac{2^x}{e^x}=0 }\), and therefore, since \(\color{black}{\displaystyle \frac{10x}{e^x}<\frac{2^x}{e^x} }\), we can note that \(\color{black}{\displaystyle \lim_{x\to\infty}\frac{10x}{e^x}=0 }\). (This is without L'H'S, although with it, it would be a matter of seconds to find the limit.)

Answer 9

What is g(x) in c?

Answer 10

why did you divide for 10x/e^x

Answer 11

isn't it multiplication

Answer 12

\(\color{black}{\displaystyle (10x)(e^{-x})=\frac{10x}{e^x} }\)

Answer 13

Just algebra, friend .... :)

Answer 14

oh because the its a negative exponent?

Answer 15

Yes:)

Answer 16

(I would advise, though, that later, when you get some free time read about L'Hospital's rule, and the indeterminate forms.)

Answer 17

Can you set up the limit for (b)?

Answer 18

yes... so lim as x-> infinity is x^2/e^4x ?

Answer 19

Yes.

Answer 20

\(\color{black}{\displaystyle \lim_{x\to\infty}\frac{x^2}{e^{4x}} }\)

Answer 21

Can you solve this limit?

Answer 22

OH i think i just figured it out

Answer 23

(If you really want to avoid L'Hospital's rule, you can either do the numerical approach - plug in bigger and bigger values, or you can do the similar thing to what I did before) ...

Answer 24

Figured out, ok ... go for it, of course:)

Answer 25

i plugged them all into a graphing calculator after i set up the limit.. and i got C (x) =(Lnx)3; g(x) = 1/x.... because this is the only one that does not approach 0

Answer 26

as it approaches infinity

Answer 27

\(\color{black}{\displaystyle \lim_{x\to \infty}\frac{(\ln x)^3}{x} }\) Like this?

Answer 28

yes thats exactly what i did

Answer 29

and then i graphed it

Answer 30

\(\small\color{black}{\displaystyle \lim_{x\to \infty}\frac{(\ln x)^3}{x}=\lim_{x\to \infty}\frac{\frac{d}{dx}(\ln x)^3}{\frac{d}{dx}x}=\lim_{x\to \infty}\frac{3(\ln x)^2\frac{1}{x}}{1}=\lim_{x\to \infty}\frac{3(\ln x)^2}{x} }\)

\(\small \color{black}{\displaystyle=\lim_{x\to \infty}\frac{\frac{d}{dx}3(\ln x)^2}{\frac{d}{dx}x}=\lim_{x\to \infty}\frac{6(\ln x)\frac{1}{x}}{1}=\lim_{x\to \infty}\frac{6\ln x}{x} =\lim_{x\to \infty}\frac{\frac{d}{dx}6\ln x}{\frac{d}{dx}x}=\lim_{x\to \infty}\frac{6\frac{1}{x}}{1}=? }\)

Answer 31

I didn't want to disappoint you, but, as I solved by L'Hospital's rule, this limit is indeed going to be 0.

Answer 32

What is f(x) in (d)?

Answer 33

Also, can you tell me the precise f and g for (a)?

Answer 34

hm thats strange because out of all the ones i graphed, that was the only one that didn't approach 0

Answer 35

@3mar

Answer 36

Well, I am here.
Sorry for being late!
I was not here!

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