Question #3f226

2 Answers
Jan 25, 2017

#lim_(x->oo) (x/(2+x))^(x-2) = 1/e^2#

Explanation:

Write the expression as:

#(x/(2+x))^(x-2) = 1/((x/(2+x))^(-(x-2))) = ((x+2)/x)^(2-x) = (1+2/x)^(2-x)#

Now take the logarithm of this expression:

#ln((x/(2+x))^(x-2)) = (2-x)ln(1+2/x) = 2ln(1+2/x)-xln(1+2/x)#

We can see that:

#lim_(x->oo) 2ln(1+2/x) = 2ln(1) = 0#

so we can ignore this term.

Focusing on the other addendum we have:

#xln(1+2/x) = 2 ln(1+2/x)/(2/x)#

Substituting #y=2/x# we have #lim_(x->oo) y(x) = 0# so that:

#lim_(x->oo) ln(1+2/x)/(2/x) = lim_(y->0) ln (1+y)/y = 1#

(you can find the explanation here )

Putting this together we can see that:

#lim_(x->oo) ln((x/(2+x))^(x-2)) = lim_(x->oo) 2ln(1+2/x)-xln(1+2/x) = 0-2*1 =-2#

Now note that:

#e^(ln((x/(2+x))^(x-2)))=(x/(2+x))^(x-2)#

so that:

#lim_(x->oo) (x/(2+x))^(x-2) = lim_(x->oo) (e^(ln((x/(2+x))^(x-2))))#

and as #e^x# is continuous in all of #RR#:

#lim_(x->oo) (x/(2+x))^(x-2) = e^(lim_(x->oo) (ln((x/(2+x))^(x-2)))) = e^(-2) = 1/e^2#

Jan 25, 2017

#"The Lim.="1/e^2.#

Explanation:

We will use the following Standard Form of Limit :

#lim_(t to oo) (1+1/t)^t=e. .......................(star)#

#"Let, "x-2=y," so, "x=2+y." Also, as "x to oo, y to oo.#

#"The Reqd. Lim., now,="lim_(y to oo) {(2+y)/(2+(2+y))}^y#

#=lim_(ytooo){(2+y)/(4+y)}^y#

#=lim_(ytooo){(y(2/y+1))/(y(4/y+1))}^y#

#lim_(ytooo){(1+2/y)/(1+4/y)}^y#

#={lim_(ytooo)(1+2/y)^y}/{lim_(ytooo)(1+4/y)^y}#

#={lim_(ytooo)(1+2/y)^(y/2)}^2/{lim_(ytooo)(1+4/y)^(y/4)}^4#

#=e^2/e^4..............[because, (star)]#

#"The Lim.="1/e^2,# as Respected Andrea S. Sir has derived.