Question

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

Answer 1

Try x=2a

Answer 2

What do you mean? Do you want me to try a variable change?

Answer 3

Yeah. Also try to use the fact that lim x->inf of (1+1/x)^x = e

Answer 4

That limit is too easy

Answer 5

Ok thank's, I got it already.

Answer 6

=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

Answer 7

set 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

Answer 8

we 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/x-1}=
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/limit-problem.pdf for few more ways of looking at it.