How to prove this $\lim$

Question

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.

Answer 1

$$ [(x+a)^n + P(x)]^{1/n} - (x + a) = (x + a)\left[ \left( 1 + \frac{P(x)}{(x+a)^n} \right)^{1/n} - 1 \right] = $$ $$ = \frac{\left[ \left( 1 + \frac{P(x)}{(x+a)^n} \right)^{1/n} - 1 \right]}{1/(x+a)} = \frac{f(x)}{g(x)} $$ 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: $$ \left( 1 + \frac{P(x)}{(x+a)^n} \right)^{\frac{1}{n} - 1} \to 1. $$

Answer 2

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 $x \to \infty$.

Answer 3

$(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} $.