# Evaluating a limit using L'Hopital

moxy

## Homework Statement

For$$Q(x) = x^k + \sum_{n=0}^{k-1} a_n x^n$$

Find $$\lim_{x→∞}\left({Q(x)^{\frac{1}{k}} - x}\right)$$

L'Hopital

## The Attempt at a Solution

$$Q(x) - x^k = \sum_{n=0}^{k-1} a_n x^n$$
$$Q(x) - x^k = (Q(x)^{\frac{1}{k}})^k - x^k = \left(Q(x)^{\frac{1}{k}} - x\right)\left([Q(x)^{\frac{1}{k}}]^{k-1} + x[Q(x)^{\frac{1}{k}}]^{k-2} + ... + x^{k-2}Q(x)^{\frac{1}{k}} + x^{k-1} \right)$$

$$Q(x)^{\frac{1}{k}} - x = \frac{Q(x) - x^k}{[Q(x)^{\frac{1}{k}}]^{k-1} + x[Q(x)^{\frac{1}{k}}]^{k-2} + ... + x^{k-2}Q(x)^{\frac{1}{k}} + x^{k-1}}$$

Then $\lim_{x→∞}{\left(Q(x)^{\frac{1}{k}} - x\right)} = \lim_{x→∞}\frac{f(x)}{g(x)}$, and if I can show $\lim_{x→∞}g(x) = ∞$, I can use L'Hopital.

Am I headed in the right direction? I can see this either working out nicely after a lot of work or being an incredible waste of time. Am I overlooking a simpler way to evaluate this limit?

## Answers and Replies

Staff Emeritus
Homework Helper
Try to factor things as follows:

$$x^k-Q(x)=x^{k-1}\left(a_{k-1}+\frac{a_{k-2}}{x}+...+\frac{a_0}{x^{k-1}}\right)$$

and

$$(Q(x)^\frac{1}{k})^{k-1}=x^{k-1}(\sqrt[k]{1+\frac{a_{k-1}}{x}+\frac{a_{k-2}}{x^2}+...+\frac{a_0}{x^{k}}})^{k-1}$$

Try to factor all terms like that and use that things like $\frac{1}{x^p}\rightarrow 0$.

moxy
Okay, so that will help me show that the limit of the denominator is ∞, but then taking the derivative is going to be a huge pain...

Staff Emeritus
Homework Helper
No, you don't need to differentiate anything!!

Do something like this

\begin{align}
\lim_{x\rightarrow +\infty}{\frac{x^3+x}{(\sqrt{x^2+x+1})^3}}
&= \lim_{x\rightarrow +\infty} {\frac{ x^3(1+\frac{1}{x^2}) }{ x^3(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}})^3} }\\
&= \lim_{x\rightarrow +\infty} {\frac{ 1+\frac{1}{x^2} }{(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}})^3} }\\
&= 1
\end{align}

moxy
Oh, okay. The problem hint suggested I use L'Hopital, so I guess I was thinking "in the box" when it came to evaluating the limit.

Thanks for the suggestions!