How Find this limits $\lim_{n\to\infty}n(x_{n}-A)$ 
let equation$$e^x+x^{2n+1}=0,n=0,1,2,\cdots$$
  I have prove this equation only have real root $x_{n}$,

show that: $\lim_{n\to \infty}x_{n}=A$ is exsit ,and Find this value,

(2):and Find
  $$\lim_{n\to\infty}n(x_{n}-A)$$

because consider equation
$$f_{n}(x)=e^x+x^{2n+1}\Longrightarrow f'(x)=e^x+(2n+1)x^{2n}>0,f(0)>0,f(-1)<0$$
and also I have prove $$-1<x_{n+1}<x_{n}<0$$
because $$f_{n+1}(x_{n})=e^{x_{n}}+(x_{n})^{2n+3}=x^{2n+1}_{n}(x^2_{n}-1)>0=f_{n+1}(x_{n+1})$$
so
$$x_{n}>x_{n+1}$$
so $$e^A+A^{2n+1}=0$$
so I can't find $A$
and the $(2)$ also can't it
 A: $$f_n(x) = e^x+x^{2n+1}$$
is an increasing function over $\mathbb{R}$. Since $f(-1)<0,f(0)=1$ the only root of $f_n(x)$ lies in $(-1,0)$.
We can refine this bound by noticing that $$ f_n\left(-1+\frac{1}{2n+1}\right)>0,$$
$$ f_n\left(-1+\frac{1}{2n+3}\right)<0,$$
hence:
$$ x_n\in\left(-1+\frac{1}{2n+3},-1+\frac{1}{2n+1}\right).$$
This just gives:
$$ A=\lim_{n\to +\infty} x_n = -1, $$
$$ \lim_{n\to +\infty}n(x_n-A) = \frac{1}{2}.$$

To prove the first inequality, we need to show that:
$$-1+\frac{1}{2n+1}>(2n+1)\log\left(1-\frac{1}{2n+1}\right),$$
that is trivial since $\frac{-\log(1-x)}{x}$ is a convex function in a neighbourhood of the origin. To prove the second inequality, we need to show that:
$$-1+\frac{1}{2n+3}<(2n+1)\log\left(1-\frac{1}{2n+3}\right),$$
or:
$$\frac{1-x}{1-2x}>\frac{-\log(1-x)}{x}$$
in a right neighbourhood of the origin. This is easily achieved by comparing the Taylor series of the LHS and the RHS.
A: This is too tricky.
The answer is $\lim_{n\to\infty} x_n=-\frac{2n+1}{2n+2}=A$.
Substitute this answer back to the equation, we have
$\lim_{n\to\infty}(e^x+x^{2n+1})=\lim_{n\to\infty}e^x+\lim_{n\to\infty}x^{2n+1}$
$=\lim_{n\to\infty}(1+\frac{x}{2n+1})^{2n+1}+\lim_{n\to\infty}x^{2n+1}$
$=\lim_{n\to\infty}(1+\frac{-\frac{2n+1}{2n+2}}{2n+1})^{2n+1}+\lim_{n\to\infty}(-\frac{2n+1}{2n+2})^{2n+1}$
$=\lim_{n\to\infty}((\frac{2n+1}{2n+2})^{2n+1}+(-\frac{2n+1}{2n+2})^{2n+1})=0$
$\lim_{n\to\infty}n(x_n-A)=\lim_{n\to\infty}n(\frac{1}{2n+2})=\frac{1}{2}-\lim_{n\to\infty}\frac{1}{2n+2}$
A: The existence of the limit $$\lim_{n\to \infty}x_{n}=A$$
has been proved above.
we solve the rest questions:
Because $$e^{x_n}=(-x_n)^{2n+1},$$
we get $$x_n=(2n+1)\log(-x_n).$$
So $$\lim_{n\to\infty}\log(-x_n)=\lim_{n\to\infty}\frac{1}{2n+1}\cdot x_n=0\cdot A=0.$$
Hence we get $$A=-1.$$
For the second one,
$$e^{x_n}=\left((1-(x_n+1))^{\frac{1}{-(x_n+1)}}\right)^{-(2n+1)(x_n+1)},$$
let $n\to\infty$, we get
$$e^{-1}=e^{\lim_{n\to\infty}-(2n+1)(x_n+1)},$$
ie.
$$\lim_{n\to\infty}-(2n+1)(x_n+1)=-1,$$
and then we will get 
$$\lim_{n\to\infty}n(x_n+1)=\frac{1}{2}.$$
