Question

help asap

Answer 1

I think it is C, and you??? from my works.

Answer 2

@3mar

Answer 3

I am on it

Answer 4

thank you.

Answer 5

Really I don't know
...Sorry for disappointing you...

Answer 6

@iwanttogotostanford.

\[\lim_{x \rightarrow \infty}g(x)*f(x) = \lim_{x \rightarrow \infty}g(x)*\lim_{x \rightarrow \infty}f(x) \]

Answer 7

answer is a if u want i can solve it

Answer 8

For the first option:
\[f(x)=10x+e^{-x}\]
\[g(x)=5x^{-1}\]
\[\lim_{x \rightarrow \infty}5x^{-1}=0\]
\[\lim_{x \rightarrow \infty}10x+e^{-x}=10x+\frac{ 1 }{ e^x }= \infty \]
\[\lim_{x \rightarrow \infty}g(x)*f(x)=\infty * 0\]
For the second option:
\[f(x)=x^2\]
\[g(x)=e^{-4x}\]
\[\lim_{x \rightarrow \infty}x^2=\infty \]
\[\lim_{x \rightarrow \infty}e^{-4x}=0 \]
\[\lim_{x \rightarrow \infty}g(x)*f(x)=\infty * 0\]
For the third option:
\[f(x)=(Lnx)^3\]
\[g(x)=x^{-1}\]
\[\lim_{x \rightarrow \infty}(Lnx)^3=\infty \]
\[\lim_{x \rightarrow \infty}x^{-1}=0 \]
\[\lim_{x \rightarrow \infty}g(x)*f(x)=\infty * 0\]
For the last option:
\[f(x)=x^{1/2}\]
\[g(x)=e^{-x}\]
\[\lim_{x \rightarrow \infty}x^{1/2}=\infty \]
\[\lim_{x \rightarrow \infty}e^{-x}=0 \]
\[\lim_{x \rightarrow \infty}g(x)*f(x)=\infty * 0\]

Answer 9

@iwanttogotostanford. think i messed up somewhere but basically.. we are finding what limits doesn't = 0 because most of these do.
checking work now

Answer 10

That leads to nothing of the choices..

Answer 11

I have checked and you are correct @sunnnystrong

Answer 12

i will post my solution for the question just give me few mins to type please

Answer 13

yeah idk D::::: lol @3mar
and i'd like to see that @hafeda

Answer 14

@sunnnystrong ugh um so confused... out of all what would you say is your best guess???

Answer 15

sure i am new to the equation system so i will need few minutes lol

Answer 16

ur answer is A btw

Answer 17

hmm gimmie me a min and it isn't a

Answer 18

@hafeda

show us please How can you reach that answer?

Answer 19

@iwanttogotostanford... D:

Answer 20

no one knows...???

Answer 21

OK WAIT MAYBE ITS A

Answer 22

I take it back lol

Answer 23

Because:

indeterminate form:
\[(\lim_{x \rightarrow \infty}x *\lim_{x \rightarrow \infty}\frac{ 1 }{ x } )\neq(\lim_{x \rightarrow \infty} (x+*\frac{ 1 }{ x })=1\]

Answer 24

\[\lim_{x \rightarrow \infty} f(x) \times g(x) \neq 0\]
\[\lim_{x \rightarrow \infty} (10x + e^-x)\times (\frac{ 1 }{5x })\]
\[\lim_{x \rightarrow \infty} x(10 + \frac{ 1 }{ e ^{x} \times x } ) ( \frac{ 1 }{ 5x } )\] ( I pulled out the x from the 10 and e)
\[\lim_{x \rightarrow \infty} (\frac{ 1 }{ 5 } ) [(\frac{ e ^{-x} }{ x }) + 10]\] ( Now the x cancel outs )
\[\frac{ 1 }{ 5 } (\lim_{x \rightarrow \infty} [\frac{ e ^{-x} }{ x }] + \lim_{x \rightarrow \infty} 10)\] (Now limit property applied to both term )
\[\frac{ 1 }{ 5 } ((\frac{ e ^{-\infty} }{ x ^{\infty} } ) + 10)\]
now i did the substitution
\[\frac{ 1 }{ 5 } ( (\frac{ 0 }{ \infty }) + 10) \]
e to the infinity is zero and zero divided by any number is zero \[\frac{ 1 }{ 5 } (0 + 10 )\]
\[\frac{ 10 }{ 5 } = 2\]

Answer 25

if u need help with any explanation i can do it no problem

Answer 26

@hafeda ah thank you i had a suspicion it was A but had no confidence- thank you sir!

Answer 27

Happy to help

Answer 28

@3mar here is the solution

Answer 29

It is A... But it isn't 2 it's 50.

Answer 30

@hafeda @sunnnystrong are you guys good at economics??? i really need help with some questions and graphs..... either or

Answer 31

https://www.wolframalpha.com/input/?i=limit+(10x%2Be%5E(-x))*+(5%2Fx)

Answer 32

according to wolfram :D

Answer 33

& no i'm pitiful sorry

Answer 34

no i am not good with graph or econ, i am sorry

Answer 35

@sunnnystrong LOL

Answer 36

aw ok do either of you know any on here that are good at it???

Answer 37

hmm on open study nah :/

@hafeda :D good job for realizing that A was an indeterminate form of infinity*0

Answer 38

Thank you for good explanation, @hafeda ... high grade technique...
I really appreciate that!

You have helped three people right here!
so Thank you very much.

Answer 39

and it is actually correct (2 not 50) @sunnnystrong

Check the given functions \(g(x)\) and \(f(x)\) again please!
https://www.wolframalpha.com/input/?i=limit+(10x%2Be%5E(-x))*+(1%2F(5x))

Answer 40

I hope you got the idea! @iwanttogotostanford

Answer 41

@3mar oh wow haha I evaluated the whole thing wrong in the first place ^^^ if you look set my work I did 5/x vs 1/5x D: ah okay I'm done lol.


If it was 5/x definitely wouldn't be a sooo XD