A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Hi, could anyone please help me with this limit? I know the answer's e^2 but I don't understand why:
It is the limit as x goes to infinity of:
(1+2/x)^x
anonymous
 5 years ago
Hi, could anyone please help me with this limit? I know the answer's e^2 but I don't understand why: It is the limit as x goes to infinity of: (1+2/x)^x

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you mean? Do you want me to try a variable change?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. Also try to use the fact that lim x>inf of (1+1/x)^x = e

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That limit is too easy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok thank's, I got it already.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0=e^ln((1+2/x)^x); xln(1+2/x)=ln(1+2/x)/(1/x);when x goes to infinity,it is 0/0; use L'Hospital's Rule, you will get it's goes to 2; so the ans is e^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0set y= limit (1+2_x)^x ln both sides ln y= limit ln(1+2/x)^x brinf exponent down limit x ln ( 1+2/x) now try to use l hospitals this way 1/x ln (1+2/x) take the derivative whish gets you 2... so ln y=2 its equals to y=e^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we can use \[1^{\infty}\]form for this limit. your problem is of the form lim {f(x)}^g(x) Now if f(inf) =1 and g(inf)=inf, then lim {f(x)}^(g(x) is given by e ^ {lim g(x){f(x)1}}. here f(x) =1+2/x and g(x)=x. also, f(inf)=1+2/inf =1+0 =1 and g(inf)=inf, thus we have e^{lim x.{1+2/x1}= e^{lim x.{2/x}=e^{lim 2} as x tends to inf leading to e^2. ALSO, visit http://sun1.sjfc.edu/~wildenbe/real_analysis/limitproblem.pdf for few more ways of looking at it.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.