Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

lim x->infinity (e^x+x)^(2/x)

ganeshie8 (ganeshie8):

Hint : factor out \(e^x\)

ganeshie8 (ganeshie8):

that may not be a very useful hint, one sec..

OpenStudy (accessdenied):

I was thinking maybe manipulating it into an indeterminate form for l'hopitals rule \( (e^x + x)^{2/x}= e^{\ln (e^x+x)^{2/x}} =e^{2/x \cdot \ln (e^x + x) }\) and taking the limit would mean taking the limit of that power, which needs l'hopital's rule

ganeshie8 (ganeshie8):

nice thats working out smoothly ! xD

OpenStudy (anonymous):

Isn't this like infinity^0

OpenStudy (anonymous):

I think it's 2. Maybe...

OpenStudy (anonymous):

It's actually \(\mathrm{e}^2\)

ganeshie8 (ganeshie8):

yes \(\lim f(x_n) = f(\lim x_n)\) as e^x is a continuous function

OpenStudy (anonymous):

Yeah e^2 seems more legit Also the graph agrees

ganeshie8 (ganeshie8):

\[\lim\limits_{x\to \infty} \left(e^x+x\right)^{2/x} = e^{\lim\limits_{x\to \infty} ~\ln \left(e^x+x\right)^{2/x} }\]

OpenStudy (anonymous):

Oh right lnL=2 so L=e^2

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!