## claudiag Group Title What is the limit as x goes to infinity of (1+2/x)^x and how do I get it? 3 years ago 3 years ago

1. dumbcow

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

2. claudiag

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?

3. claudiag

through* sorry

4. dumbcow

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

5. dumbcow

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 $e ^{\lim_{x \rightarrow \infty}\frac{\frac{1}{x+2}-\frac{1}{x}}{-\frac{1}{x ^{2}}}}$

6. dumbcow

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

7. claudiag

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!!