Question

Question #3f226

Answer 1

`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`

Answer 2

`"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.