Question

lim as x approaches positive infinity of (1 + 1/4x) ^ 3x

Answer 1

I would use the fact that e^(ln(y))=y

Answer 2

\[\lim_{x \rightarrow \infty}e^{3x \ln(1+\frac{1}{4x})}=\lim_{x \rightarrow \infty}e^{3 \frac{\ln(1+\frac{1}{4x})}{\frac{1}{x}}}\]

Answer 3

now use l'hospitl rule since you have 0/0 above in the exponent

Answer 4

so the answer would be e^3

Answer 5

I don't think that is correct....

Answer 6

what is the derivative of ln(1+1/(4x)) and what is the derivative of 1/x?

Answer 7

(1/1+1/4x) * -1/x^2 and -1/x^2

Answer 8

see that is what is going on you are forgetting the constant multiple inside

Answer 9

\[\frac{d}{dx}\ln(1+\frac{1}{4x})=\frac{-1}{4x^2} \frac{1}{1+\frac{1}{4x}} \\ \frac{d}{dx}\frac{1}{x}=\frac{-1}{x^2}\]

Answer 10

\[\lim_{x \rightarrow \infty}e^{3x \ln(1+\frac{1}{4x})}=\lim_{x \rightarrow \infty}e^{3 \frac{\ln(1+\frac{1}{4x})}{\frac{1}{x}}} =\lim_{x \rightarrow \infty}e^{3 \frac{-1}{4x^2} \frac{1}{1+\frac{1}{4x}} \cdot (-x^2)}\]

Answer 11

so it would result in e^3/4

Answer 12

yes

Answer 13

awesome!! thank you for your help!!

Answer 14

the x^2's will cancel
\[\lim_{x \rightarrow \infty}e^{3x \ln(1+\frac{1}{4x})}=\lim_{x \rightarrow \infty}e^{3 \frac{\ln(1+\frac{1}{4x})}{\frac{1}{x}}} =\lim_{x \rightarrow \infty}e^{3 \frac{1}{4\cancel{-x^2}} \frac{1}{1+\frac{1}{4x}} \cdot (\cancel{-x^2})} \\ =\lim_{x \rightarrow \infty}e^{\frac{3}{4} \frac{1}{1+\frac{1}{4x}}}\]
and then that one part goes to 1/(1+0)=1
and then you have 3/4*1=3/4
and blah blah

Answer 15

or in the words of Seinfeld
yada yada yada

Answer 16

or if you're allowerd, you could also use the standard limit \(\large\lim\limits_{n\to\infty} (1+x/n)^n = e^x \)

Answer 17

freckles work is a nice derivation of above standard result i think..