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((1+x)/(2+x))^(1+x)

Limit of the function ((1+x)/(2+x))^(1+x)

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            1 + x
     /1 + x\     
 lim |-----|     
x->oo\2 + x/     
limx(x+1x+2)x+1\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1}
Limit(((1 + x)/(2 + x))^(1 + x), x, oo, dir='-')
Detail solution
Let's take the limit
limx(x+1x+2)x+1\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1}
transform
limx(x+1x+2)x+1\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1}
=
limx((x+2)1x+2)x+1\lim_{x \to \infty} \left(\frac{\left(x + 2\right) - 1}{x + 2}\right)^{x + 1}
=
limx(1x+2+x+2x+2)x+1\lim_{x \to \infty} \left(- \frac{1}{x + 2} + \frac{x + 2}{x + 2}\right)^{x + 1}
=
limx(11x+2)x+1\lim_{x \to \infty} \left(1 - \frac{1}{x + 2}\right)^{x + 1}
=
do replacement
u=x+21u = \frac{x + 2}{-1}
then
limx(11x+2)x+1\lim_{x \to \infty} \left(1 - \frac{1}{x + 2}\right)^{x + 1} =
=
limu(1+1u)u1\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u - 1}
=
limu((1+1u)u1+1u)\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- u}}{1 + \frac{1}{u}}\right)
=
limu11+1ulimu(1+1u)u\lim_{u \to \infty} \frac{1}{1 + \frac{1}{u}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}
=
limu(1+1u)u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}
=
((limu(1+1u)u))1\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1}
The limit
limu(1+1u)u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}
is second remarkable limit, is equal to e ~ 2.718281828459045
then
((limu(1+1u)u))1=e1\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1} = e^{-1}

The final answer:
limx(x+1x+2)x+1=e1\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1} = e^{-1}
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
02468-8-6-4-2-10100.01.0
Rapid solution [src]
 -1
e  
e1e^{-1}
The graph
Limit of the function ((1+x)/(2+x))^(1+x)