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claudiag
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
What do you mean? Do you want me to try a variable change?
Yeah. Also try to use the fact that lim x->inf of (1+1/x)^x = e
Ok thank's, I got it already.
=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
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
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.