Question

Can't Solve!! Been at this for hours! ILL GIVE YOU A MEDAL IF YOU CAN SOLVE THIS. Limit as x approaches infinity of (1+3/x)^(x/15) [This involves using l'hospital's rule and indeterminate forms]

Answer 1

\[e^{\ln((1+\frac{3}{x})^\frac{x}{15})}=(1+\frac{3}{x})^\frac{x}{15}\]
use this e^ln thing

Answer 2

I know that much. But try and solve for it

Answer 3

Youll see what I mean when you try and solve

Answer 4

\[\lim_{x \rightarrow \infty} \frac{x}{15} \ln(1+\frac{3}{x})=\lim_{x \rightarrow \infty} \frac{\ln(1+\frac{3}{x})}{\frac{15}{x}}\]

Answer 5

we will come back to the base e thing here in a bit

Answer 6

But here you have 0/0 as x-infinity

Answer 7

use l'hosptal's rule

Answer 8

@Dust_60 do you understand ?

Answer 9

can you differentiate the numerator and denominator?

Answer 10

wait hold on

Answer 11

Ok so this is what I get after I differentiate and simplify \[\frac{ x^2 }{ -15+\frac{ -45 }{ x } }\]

Answer 12

\[\frac{d}{dx}\ln(1+\frac{3}{x})=\frac{0-\frac{3}{x^2}}{1+\frac{3}{x}}=\frac{\frac{-3}{x^2}}{1+\frac{3}{x}} \\ \frac{d}{dx}(\frac{15}{x})=-\frac{15}{x^2}\]

Answer 13

\[\frac{ \frac{d}{dx} (\ln(1+\frac{3}{x}))}{\frac{d}{dx} (\frac{15}{x})} =\frac{-\frac{3}{x^2}}{1+\frac{3}{x}} \cdot \frac{-x^2}{15}\]

Answer 14

Those x^2's should cancel there

Answer 15

Oh I see I forgot to take the derivative of g(x) forgot about chain rule lol

Answer 16

\[\frac{ \frac{d}{dx} (\ln(1+\frac{3}{x}))}{\frac{d}{dx} (\frac{15}{x})} =\frac{-\frac{3}{ \cancel{x^2}}}{1+\frac{3}{x}} \cdot \frac{-\cancel{x^2}}{15}\]

Answer 17

hmm so where do you go from there?

Answer 18

\[\lim_{x \rightarrow \infty} \frac{x}{15} \ln(1+\frac{3}{x})=\lim_{x \rightarrow \infty} \frac{\ln(1+\frac{3}{x})}{\frac{15}{x}}=\lim_{x \rightarrow \infty}\frac{-3}{-15(1+\frac{3}{x})} \]

Answer 19

Now -3/-15 can be reduced to 1/5

Answer 20

but where does 1/(1+3/x) go as x goes to infinity

Answer 21

Your good......

Answer 22

Thanks so much

Answer 23

Ok I guess you got it from here?

Answer 24

yea it should be e^(1/5) or the 5th root of e

Answer 25

sounds good