Question

What is the limit as x goes to infinity of (1+2/x)^x and how do I get it?

Answer 1

put in indeterminate form then apply L'Hopitals rule by differentiating top and bottom
rewrite it as
\[\frac{(x+2)^x}{x^x}\]

Answer 2

I think I get it but, the derivation is kind of weird... or I don't really know if I'm doing it right... would it bother you too much to walk me throught it?

Answer 3

through* sorry

Answer 4

\[y = x^{x} \rightarrow \ln y= x \ln x\]
Now differentiate both sides
\[\frac{1}{y}\frac{dy}{dx} = \ln x +1\]
multiply by y on both sides where y = x^x
\[\frac{dy}{dx} = x^{x}(\ln x+1)\]

Answer 5

ok i realized i was wrong in my approach, those derivatives do get messy and dont help find a solution.
my apologies, here is the correct solution step by step
transform problem using natural log identity, e^ln(x) = x
\[e ^{\lim_{x \rightarrow \infty}\ln (\frac{x+2}{x})^{x}}\]
use log rules to expand and move x to denominator
\[e ^{\lim_{x \rightarrow \infty}\frac{\ln (x+2) -\ln x}{\frac{1}{x}}}\]
here we notice if we evaluate limit we get inf-inf/0 which is indeterminate so apply L'hopitals rule and differentiate top and bottom
\[e ^{\lim_{x \rightarrow \infty}\frac{\frac{1}{x+2}-\frac{1}{x}}{-\frac{1}{x ^{2}}}}\]

Answer 6

combine fractions on top then multiply by -x^2 on top and bottom
\[e ^{\lim_{x \rightarrow \infty}\frac{2x}{x+2}}\]
if you evaluate again you'll get infinity over infinity so differentiate again
\[e ^{\lim_{x \rightarrow \infty}2}\rightarrow e ^{2}\]
so the limit of the function is e^2

Answer 7

Thank's I'll try it out. I don't really understand the first step in which you transform the problem using the idientity, I don't understand how is that possible. But I do understand the resto of it so I'll find my teacher and ask him about that. Thanks a lot!!