Evaluating limit making it $\frac{\infty}{\infty}$ and using L'Hopital Rule Let $P(x)=x^n+\displaystyle\sum\limits_{k=0}^{n-1}a_kx^k$. Find $$ \lim_{x \to +\infty} ([P(x)]^{1/n}-x) $$
I know that in order to solve this problem I need to multiply it by something that will make it $\frac{\infty}{\infty}$ and then use L'Hopital Rule. I also know that the answer should be $1/n$ if I am not mistaken. I have tried multiplying the expression by $\frac{e^x}{e^x}$ and then using L'H. Rule but with not much success. Any suggestions on how I should proceed? Thank you for the help. 
 A: An idea:
$$P(x)^{1/n}-x=x\left[\left(1+\frac{a_0}{x^n}+\ldots+\frac{a_{n-1}}x\right)^{\frac1n}-1\right]=\frac{\left(1+\frac{a_0}{x^n}+\ldots+\frac{a_{n-1}}x\right)^{\frac1n}-1}{\frac1x}$$
Now you can use l'Hospital (can you see why?) and get that your limit equals
$$\lim_{x\to\infty}\frac{\frac1n\left(-\frac{na_0}{x^{n+1}}-\ldots-\frac{a_{n-1}}{x^2}\right)\left(1+\frac{a_0}{x^n}+\ldots+\frac{a_{n-1}}x\right)^{\frac1n-1}}{-\frac1{x^2}}=$$
$$=\frac1n\;\lim_{x\to\infty}\,\left(\frac{na_0}{x^{n-1}}+\ldots+\frac{2a_{n-2}}x+a_{n-1}\right)\left(1+\frac{a_0}{x^n}+\ldots+\frac{a_{n-1}}x\right)^{\frac1n-1}=\frac{a_{n-1}}n$$
A: Here is a way of answering this question without appealing to l'Hospital's rule.  In principle, any limit involving radicals and rational functions can be resolved by algebraic manipulations.  In fact, it's the same idea that went into the answers provided by DonAntonio and Cronus.
We wish to find the limit: $$\lim_{x\to\infty} (P(x)^{1/n} - x)$$ where $P(x) = x^n + \sum_{k=0}^{n-1} a_k x^k$ is a polynomial of degree $n$.
First we remember the polynomial equation: $$(y-x)(x^{n-1}y^0 + x^{n-2}y^1 + \cdots + x^0y^{n-1}) = y^n - x^n$$
Now take $y = P(x)^{1/n}$.  This enables us to rewrite the limit as:
$$\begin{array}{rcl}\lim_{x\to\infty} (P(x)^{1/n} - x) &=& \lim_{x\to\infty} \frac{P(x) - x^n}{(x^{n-1}(P(x)^{1/n})^0 + x^{n-2}(P(x)^{1/n})^1 + \cdots + x^0(P(x)^{1/n})^{n-1})}\\
&=& \lim_{x\to\infty} \frac{\sum_{k=0}^{n-1} a_k x^k}{(x^{n-1}(P(x)^{1/n})^0 + x^{n-2}(P(x)^{1/n})^1 + \cdots + x^0(P(x)^{1/n})^{n-1})}\end{array}$$
Now the key observation is that we can view every term in the denominator as being $x^{n-1}$ in a limiting sense:
$$\begin{array}{rcl}x^{n-k}(P(x)^{1/n})^{k-1}&=&x^{n-k}((x^n + a_{n-1} x^{n-1} + \cdots)^{1/n})^{k-1}\\
&=&x^{n-k}(x (1 + a_{n-1} x^{-1} \cdots )^{1/n})^{k-1}\\
&=&x^{n-1}(1 + a_{n-1} x^{-1} \cdots )^{(k-1)/n}\end{array}$$
Note that $\lim_{x\to\infty} (1 + a_{n-1} x^{-1} \cdots )^{(k-1)/n} = 1$.  For convenience, note that we can write these terms as: $$x^{n-1}(P(x)/x^n)^{(k-1)/n}$$
Now let us return the the limit we are concerned with:
$$\begin{array}{rcl}\lim_{x\to\infty} (P(x)^{1/n} - x) &=& \lim_{x\to\infty} \frac{\sum_{k=0}^{n-1} a_k x^k}{(x^{n-1}(P(x)^{1/n})^0 + x^{n-2}(P(x)^{1/n})^1 + \cdots + x^0(P(x)^{1/n})^{n-1})}\\
&=&\lim_{x\to\infty} \frac{\sum_{k=0}^{n-1} a_k x^k}{x^{n-1}(((P(x)/x^n)^{1/n})^0 + ((P(x)/x^n)^{1/n})^1 + \cdots + ((P(x)/x^n)^{1/n})^{n-1})}\\
&=&\lim_{x\to\infty} \frac{a_{n-1} + a_{n-2} x^{-1} + \cdots + a_0 x^{-(n-1)}}{((P(x)/x^n)^{1/n})^0 + ((P(x)/x^n)^{1/n})^1 + \cdots + ((P(x)/x^n)^{1/n})^{n-1}}\end{array}$$
Finally notice this is simply a combination of limits we already know.  The numerator tends to $a_{n-1}$ as $x \to \infty$, and each of the terms in the denominator (of which there are $n$ terms) tends to 1 as we already stated.
Thus,
$$\lim_{x\to\infty} (P(x)^{1/n} - x) = \frac{a_{n-1}}{n}.$$
A: Hint:
$$P(x)^{\frac{1}{n}} = x\Big(1 + \sum_{k = 0}^{n - 1}a_kx^{k-n}\Big)^{\frac{1}{n}}$$ $$P(x)^{\frac{1}{n}} - x = \frac{\Big(1 + \sum_{k = 0}^{n - 1}a_kx^{k-n}\Big)^{\frac{1}{n}} - 1}{\frac{1}{x}}$$ Now you can use L'Hopital rule!
A: Compute the Taylor series of P(x)^(1/n)-x around x = Infinity. The result given by DonAntonio is almost immediate.
