Question

limx->infinity of (1+a/x)^bx

Please look at my work below and tell me were I am messing up! @Calculus1

Answer 1

Intermediate form type 1^infinity
lny=limx->infinity (bx)ln(1+(a/x))

Answer 2

Intermediate form type infinity * 0

lny = limx->infinity ln(1+(a/x))/(1/bx)

or
ln(1+(a/x)/(-1bx^-2)

Answer 3

Assuming that means
\[\frac{\ln(1+\frac{a}{x})}{\frac{1}{bx^2}} \]
good so far.

Answer 4

Not x^2 in the bottom, sorry...

Answer 5

\[e^{ab}\] is my guess. is it right?

Answer 6

^ yup

Answer 7

wait, i'm confused, give me a second to think

Answer 8

Okay so I'm taking derivative of
lny= limx->infinity ln(1+a/x)/bx =

1/(1+a/x)/-1(bx)^-2

Answer 9

\[1/(1+a/x)/-bx^{-2}\]

Answer 10

And this is where I am stuck. Do I plug in infinity now?

Answer 11

Not quite. There's a mistake somewhere, it should be
\[ \ln(y) = bx \ln(1+\frac{a}{x}) =\frac{\ln(1+\frac{a}{x})}{\frac{1}{bx}}\]

Answer 12

You need to do l'hospitals now though! I got that^

Answer 13

????/

Answer 14

\[ \lim_{x \rightarrow \infty} \frac{\ln(1+\frac{a}{x})}{\frac{1}{bx}} = \lim_{x \rightarrow \infty} \frac{\frac{\frac{-a}{x^2}}{1+\frac{a}{x}}}{\frac{-1}{bx^2}} = \lim_{x \rightarrow \infty} \frac{ab}{1+\frac{a}{x}}=ab\]

Answer 15

forgot about that chain rule on a/x, damn this is getting messy

Answer 16

Therefore,
\[\lim_{x \rightarrow \infty} (1+\frac{a}{x})^{bx} = e^{ab}\]
FYI those of us who can and wish to take the time type in LaTeX on here, which makes our answers somewhat slow but infinitely more readable than regular plain text, so if I or somebody else is typing for awhile, we're just doing this:

"lim_{x \rightarrow \infty} (1+\frac{a}{x})^{bx} = e^{ab}"

which shows up like this

\[lim_{x \rightarrow \infty} (1+\frac{a}{x})^{bx} = e^{ab}\]

Answer 17

how did you get ab for that ugly numerator?

Answer 18

you should really do this question with your eyeballs.

Answer 19

\[\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n=e\] and
\[\lim_{n\rightarrow \infty}(1+\frac{x}{n})^n=e^x\]

Answer 20

What, you mean look it up?

Answer 21

What??

Answer 22

I don't understand what you did from this:

((-1/x^2)/(1+a/x))/(-1/bx^2)

Answer 23

Do you know how to simplify improper fractions? Multiply the top and bottom by -bx^2.

Answer 24

like, that next step. Which you somehow got ab/1+ax

Answer 25

Erm, how did you multiply ((-a/x^2)/1+a/x)) by -bx^2??

Answer 26

Um,
\[\frac{\frac{-a}{x^2}*-bx^2}{1+\frac{a}{x}} \]

Answer 27

Okay, that made it clear^^

Thank you so much. And I'm really bad at simplifying improper fractions.