How to prove this $\lim$ Let $n$ be integer, $n\geq 2$, and $P(x)$ be a polynomial with degree < $n-1$.
How to prove that $$\lim_{x\to\infty}\left(\sqrt[n]{(x+a)^n+P(x)}-x-a\right)=0$$
I try to use conjugate but it is very difficult.
Thank you for helping.  
 A: $$
[(x+a)^n + P(x)]^{1/n} - (x + a) = (x + a)\left[
\left(
1 + \frac{P(x)}{(x+a)^n}
\right)^{1/n} - 1
\right] =
$$
$$
= \frac{\left[
\left(
1 + \frac{P(x)}{(x+a)^n}
\right)^{1/n} - 1
\right]}{1/(x+a)} = \frac{f(x)}{g(x)}
$$
and you may use L'Hospital's rule.
$$
g'(x) = -\frac{1}{(x+a)^2},
$$
$$
f'(x) = \frac{1}{n}\left(
1 + \frac{P(x)}{(x+a)^n}
\right)^{\frac{1}{n} - 1}
\frac{P'(x)(x+a)^n - P(x)n(x+a)^{n-1}}{(x+a)^{2n}}
$$
One can show that $$\lim_{x\to\infty}\frac{f'(x)}{g'(x)} = 0$$ so
$$\lim_{x\to\infty}\frac{f(x)}{g(x)} = 0.$$
Note:
$$
\left(
1 + \frac{P(x)}{(x+a)^n}
\right)^{\frac{1}{n} - 1} \to 1.
$$
A: Let $$A = \sqrt[n]{(x + a)^{n} + P(x)}, B = (x + a), C = \frac{P(x)}{(x + a)^{n}}, D = \frac{P(x)}{(x + a)^{n - 1}}$$ and then we can see that \begin{align}
A - B &= \frac{A^{n} - B^{n}}{A^{n - 1} + A^{n - 2}B + \dots + AB^{n - 2} + B^{n - 1}}\notag\\
&= \dfrac{P(x)}{(x + a)^{n - 1}\left\{\left(1 + \dfrac{P(x)}{(x + a)^{n}}\right)^{(n - 1)/n} + \left(1 + \dfrac{P(x)}{(x + a)^{n}}\right)^{(n - 2)/n} + \cdots + 1\right\}}\notag\\
&= \frac{D}{(1 + C)^{(n - 1)/n} + (1 + C)^{(n - 2)/n} + \cdots + 1}\tag{1}
\end{align} Note that $P(x)$ is a polynomial of degree less than $(n - 1)$ and hence $C, D$ both tend to $0$ as $x \to \infty$. Clearly in the equation $(1)$ we can see that the numerator tends to $0$ and denominator tends to $1 + 1 + \dots + 1 = n$ so that $(A - B) \to 0$ as $x \to \infty$.
A: $(x+a)^n + P(x)
=x^n + nax^{n-1} + O(x^{n-2}) + P(x)
=x^n + nax^{n-1} + O(x^{n-2})
$
since
$P(x)
=O(x^{n-2})
$.
Therefore
$\begin{array}\\
((x+a)^n + P(x))^{1/n}
&=(x^n + nax^{n-1} + O(x^{n-2}))^{1/n}\\
&=x(1 + na/x + O(x^{-2}))^{1/n}\\
&=x(1 + a/x + O(x^{-2}))\\
&=x+a + O(1/x)\\
\end{array}
$.
