Question

lim as x approaches positive infinity (1-3/x)^x

i get log2 but my book say the answer is e^-3

Answer 1

how do you get that?

Answer 2

\[\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^{nt}=e^{rt}\]

if you let r=-3 and t=1 you get the result

Answer 3

http://en.wikipedia.org/wiki/Exponential_function#Formal_definition

Answer 4

what happen to the x exponent

Answer 5

x is a dummy variable. call it what you want

still the same result

\[\lim_{x\to\infty}\left(1+\frac{r}{x}\right)^{xt}=e^{rt}\]

Answer 6

no i understand that im talking about the original eqution how do you know which is which. like how do you get that from the orginal

Answer 7

The limit cwrw238 and Zarkon are using is the precise definition of the number \(e\approx2.718\).

Answer 8

Please help

Answer 9

this would help https://answers.yahoo.com/question/index?qid=20100123142609AATKWxy

Answer 10

just follow the steps :)

Answer 11

@eggshell You requested my help. Do you still need help?

Answer 12

@eggshell yeah do you :) ?

Answer 13

yes im looking at the link that was posted and im confused on what happens to the natural log and how it just jumps to -1-3 over x^2. its the picture o

Answer 14

so you know e^(ln(y))=y

Answer 15

we are going to ignore the e part until later

Answer 16

lets just look at \[\lim_{x \rightarrow \infty}\ln[(1-\frac{3}{x})^x]\]

Answer 17

so you know by property of log we can bring down that power in the inside there

Answer 18

yes

Answer 19

explained here : http://www.wolframalpha.com/input/?i=%28%281-3%29%2Fx%29%5Ex

Answer 20

\[\lim_{x \rightarrow \infty}x \ln(1-\frac{3}{x})=\lim_{x \rightarrow \infty}\frac{\ln(1-\frac{3}{x})}{\frac{1}{x}}\]
here we have 0/0

Answer 21

we need to use l'hosptal's rule

Answer 22

first what is the derivative of ln(1-3/x)?

Answer 23

and then what is the derivative of 1/x?

Answer 24

\[\lim_{x \rightarrow \infty} \frac{(\ln(1-\frac{3}{x}))' }{(\frac{1}{x})'}=\]

Answer 25

1/(1-3/x) and x?

Answer 26

\[(\ln(1-\frac{3}{x}))'=\frac{(1-\frac{3}{x})'}{1-\frac{3}{x}}=\frac{0+\frac{3}{x^2}}{1-\frac{3}{x}}=\frac{\frac{3}{x^2}}{1-\frac{3}{x}}=\frac{3}{x^2-3x}\]

Answer 27

the derivative of 1/x is not x

Answer 28

\[(\frac{1}{x})'=(x^{-1})'=-1x^{-1-1}=-1x^{-2}=\frac{-1}{x^2}\]

Answer 29

\[\lim_{x \rightarrow \infty} \frac{(\ln(1-\frac{3}{x}))' }{(\frac{1}{x})'}=\lim_{x \rightarrow \infty}\frac{3}{x^2-3x} \cdot (-x^2)\]

Answer 30

see if you can find the limit now

Answer 31

i get zero

Answer 32

No do it again

Answer 33

you have -3x^2/(x^2-3x)
it is not zero

Answer 34

to he limit I plug in zero that's all I get

Answer 35

why plug in 0?

Answer 36

also if you plug in 0 we get 0/0
but we shouldn't be pluggin in 0 anyways

Answer 37

we should know from earlier in calculus that 1/x^n approaches 0 as x approach infinity for n greater than or equal to 1

Answer 38

why not I thought that's what we did with infinity

Answer 39

\[\lim_{x \rightarrow \infty}\frac{-3x^2}{x^2-3x} \ \lim_{x \rightarrow \infty}\frac{-3}{1-\frac{1}{x}}=\frac{\lim_{x \rightarrow \infty}(-3)}{\lim_{x \rightarrow \infty}(1)-\lim_{x \rightarrow \infty}( \frac{1}{x})}\]

Answer 40

though you didn't have to do that either..
you could have used l'hosptal on the -3x^2/(x^2-3x) thing

Answer 41

since you have pm infinity/ininifty

Answer 42

before doing that you could also cancel a factor of x

Answer 43

and still use l'hospital lol

Answer 44

but yeah you don't plug in 0 when you have x approaches something that isn't 0

Answer 45

so I just cancel out? my teacher told me to plus ito plug in zero with infintiy

Answer 46

cancel out?

Answer 47

\[\lim_{x \rightarrow \infty}\frac{-3x^2}{x^2-3x} \ \lim_{x \rightarrow \infty}\frac{-3}{1-\frac{1}{x}}=\frac{\lim_{x \rightarrow \infty}(-3)}{\lim_{x \rightarrow \infty}(1)-\lim_{x \rightarrow \infty}( \frac{1}{x})} \\ =\frac{-3}{1-0}=\frac{-3}{1}=-3\]